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Square Roots and 2nd Degree Equations

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Radicals and Square-Equations
Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it.
Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = c.–3 = Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = 1.732.. (calculator) We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = 1.732.. (calculator) Note that the square of both +3 and –3 is 9, but we designate sqrt(9) or οƒ–9 to be +3. We say β€œβ€“3” is the β€œnegative of the square root of 9”. We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Since 30 is about half way between 25 and 36, Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 0 = 0 1 12 = 1 1 = 1 2 22 = 4 4 = 2 3 32 = 9 9 = 3 4 42 = 16 16 = 4 5 52 = 25 25 = 5 6 62 = 36 36 = 6 7 72 = 49 49 = 7 8 82 = 64 64 = 8 9 92 = 81 81 = 9 10 102 = 100 100 = 10 11 112 = 121 121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence 25 < 30 <36 or 5 < 30 < 6 Since 30 is about half way between 25 and 36, so we estimate that30  5.5. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Since 30 is about half way between 25 and 36, so we estimate thatοƒ–30 ο‚» 5.5. In fact οƒ–30 ο‚» 5.47722…. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator exact answer approximate answer (If c<0, there is no real solution.)
Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator exact answer approximate answer Square-roots numbers show up in geometry for measuring distances because of the Pythagorean Theorem. (If c<0, there is no real solution.)
Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0.
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0.
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Note that οƒ–x Β± y β‰  οƒ–x Β± οƒ–y Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2
Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Note that οƒ–x Β± y β‰  οƒ–x Β± οƒ–y so that οƒ–13 = οƒ–4 + 9 β‰  οƒ–4 + οƒ–9 = 5 οƒ–x2 – 4 β‰  x – 2 Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2 where as οƒ–(x – 2)2 = οƒ–x2 – 4x + 4 = l x – 2 l
Solving 2nd Degree Equations ax2 + bx + c = 0
Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.
Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers approx. answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF). In this course, all approx. answers are given with three significant digits, i.e. starting from the first nonzero digit, round off the answer to a three digit number.
2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0 2x(x – 4) = 0
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3, x = 2 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor and extract the solutions
Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3, x = 2 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor and extract the solutions If the equation is not easily factorable, use the formula below.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x =
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots,
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
–bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 ο‚» {1.22 –0.549 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0
Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20
Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20 and –20 does not exist. Hence it does not have any real number solution.
Rules of Radicals Exercise 3. A. Solve for x. Give both the exact and approximate answers. If the answer does not exist, state so. 1. x2 = 1 2. x2 – 5 = 4 3. x2 + 5 = 4 4. 2x2 = 31 5. 4x2 – 5 = 4 6. 5 = 3x2 + 1 7. 4x2 = 1 8. x2 – 32 = 42 9. x2 + 62 = 102 10. 2x2 + 7 = 11 11. 2x2 – 5 = 6 12. 4 = 3x2 + 5 B. Solve the following equations by factoring. 5. x2 – 3x = 10 9. x3 – 2x2 = 0 6. x2 = 4 7. 2x(x – 3) + 4 = 2x – 4 10. 2x2(x – 3) = –4x 8. x(x – 3) + x + 6 = 2x2 + 3x 1. x2 – 3x – 4 = 0 2. x2 – 2x – 15 = 0 3. x2 + 7x + 12 = 0 4. –x2 – 2x + 8 = 0 11. 4x2 = x4 12. 7x2 = –4x3 – 3x 13. 5 = (x + 2)(2x + 1) 14. (x + 1)2 = x2 + (x – 1)2 15. (x + 3)2 – (x + 2)2 = (x + 1)2
C. Solve the following equations by the quadratic formula. If the answers are not real numbers, just state so. 1. x2 – x + 1 = 0 2. x2 – x – 1 = 0 3. x2 – 3x – 2 = 0 4. x2 – 2x + 3 = 0 5. 2x2 – 3x – 1 = 0 6. 3x2 = 2x + 3 Equations
(Answers to odd problems) Exercise C. 1. No real solution 3. π‘₯ = 3 2 Β± 17 2 5. π‘₯ = 1 4 3 Β± 17 1. π‘₯ = βˆ’1, π‘₯ = 4 3. π‘₯ = βˆ’4, π‘₯ = βˆ’3 5. π‘₯ = βˆ’2, π‘₯ = 5 7. π‘₯ = 2 9. π‘₯ = 0, π‘₯ = 2 11. π‘₯ = Β±2, π‘₯ = 0 13. π‘₯ = βˆ’3, π‘₯ = 1 2 15. π‘₯ = 2(2 Β± 5) Equations 1. x = Β±1 3. Doesn’t exist. 7. x = Β±1/2 9. x = Β±8 11. x = ±√11/2 Exercise A. 5. x = Β±3/2 Exercise B.

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3 square roots and 2nd degree equations nat-e

  • 2. Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it.
  • 3. Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 4. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Radicals and Square-Equations We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 5. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 6. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = c.–3 = Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 7. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 8. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 9. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = 1.732.. (calculator) We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 10. β€œ9 is the square of 3” may be rephrased backwards as β€œ3 is the square root of 9”. Example D. a. Sqrt(16) = 4 c.–3 = doesn’t exist Radicals and Square-Equations Definition: If a is > 0, and a2 = x, then we say a is the square root of x. This is written as sqrt(x) = a, or οƒ–x = a. b. οƒ–1/9 = 1/3 d. οƒ–3 = 1.732.. (calculator) Note that the square of both +3 and –3 is 9, but we designate sqrt(9) or οƒ–9 to be +3. We say β€œβ€“3” is the β€œnegative of the square root of 9”. We need a scientific calculator for thin section. You may use the digital calculators on your personal devices. Make sure the calculator could displace your input so that you may check the input before executing it. Square Root
  • 11. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize.
  • 12. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 13. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 14. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 15. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 16. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 17. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Since 30 is about half way between 25 and 36, Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 18. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Since 30 is about half way between 25 and 36, so we estimate thatοƒ–30 ο‚» 5.5. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 19. 0 02 = 0 οƒ–0 = 0 1 12 = 1 οƒ–1 = 1 2 22 = 4 οƒ–4 = 2 3 32 = 9 οƒ–9 = 3 4 42 = 16 οƒ–16 = 4 5 52 = 25 οƒ–25 = 5 6 62 = 36 οƒ–36 = 6 7 72 = 49 οƒ–49 = 7 8 82 = 64 οƒ–64 = 8 9 92 = 81 οƒ–81 = 9 10 102 = 100 οƒ–100 = 10 11 112 = 121 οƒ–121 = 11 We may estimate the sqrt of other small numbers using this table. For example, 25 < 30 < 36 hence οƒ–25 < οƒ–30 <οƒ–36 or 5 < οƒ–30 < 6 Since 30 is about half way between 25 and 36, so we estimate thatοƒ–30 ο‚» 5.5. In fact οƒ–30 ο‚» 5.47722…. Radicals and Square-Equations Following are the square numbers and square-roots that one needs to memorize. These numbers are special because many mathematics exercises utilize square numbers.
  • 20. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations
  • 21. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations (If c<0, there is no real solution.)
  • 22. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 (If c<0, there is no real solution.)
  • 23. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 (If c<0, there is no real solution.)
  • 24. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 (If c<0, there is no real solution.)
  • 25. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 (If c<0, there is no real solution.)
  • 26. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. (If c<0, there is no real solution.)
  • 27. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 (If c<0, there is no real solution.)
  • 28. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 (If c<0, there is no real solution.)
  • 29. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator (If c<0, there is no real solution.)
  • 30. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator exact answer approximate answer (If c<0, there is no real solution.)
  • 31. Equations of the form x2 = c has two answers: x = +οƒ–c or –c if c>0. Radicals and Square-Equations Example E. Solve the following equations. a. x2 = 25 x = Β±οƒ–25 = Β±5 b. x2 = –4 Solution does not exist. c. x2 = 8 x = Β±οƒ–8 ο‚» Β±2.8284.. by calculator exact answer approximate answer Square-roots numbers show up in geometry for measuring distances because of the Pythagorean Theorem. (If c<0, there is no real solution.)
  • 32. Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0.
  • 33. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0.
  • 34. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1
  • 35. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1
  • 36. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2
  • 37. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2
  • 38. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2
  • 39. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Note that οƒ–x Β± y β‰  οƒ–x Β± οƒ–y Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2
  • 40. Square Rule: οƒ–x2 =οƒ–x οƒ–x = x Multiplication Rule: οƒ–xΒ·y = οƒ–xΒ·οƒ–y Rules of Radicals For the following discussion of square roots, the variables x and y are assumed β‰₯ 0. Division Rule: οƒ–x/y = οƒ–x /οƒ–y Hence οƒ–3 Β·οƒ–3 = 3, οƒ–(x + 1)2 = οƒ–x + 1 Β· οƒ– x + 1 = x + 1 Hence οƒ–12 Β·οƒ–3 = οƒ–36 = 6 οƒ–12 Β·οƒ–1/3 = οƒ–12 Β·(1/3) = οƒ–1/4 = 1/2 Note that οƒ–x Β± y β‰  οƒ–x Β± οƒ–y so that οƒ–13 = οƒ–4 + 9 β‰  οƒ–4 + οƒ–9 = 5 οƒ–x2 – 4 β‰  x – 2 Hence οƒ–12 /οƒ–3 = οƒ–12/3 = οƒ–4 = 2 where as οƒ–(x – 2)2 = οƒ–x2 – 4x + 4 = l x – 2 l
  • 41. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 42. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing.
  • 43. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials.
  • 44. Solving 2nd Degree Equations ax2 + bx + c = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 45. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 46. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 47. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 48. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 49. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 50. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 51. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 52. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d. II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 53. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 54. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF).
  • 55. Solving 2nd Degree Equations ax2 + bx + c = 0 The Square-Root Method Use the square-root method if there is no x-term. I. Solve for the x2 and get the form x2 = d II. Then x = Β±οƒ–d are the roots. Example F. Solve 3x2 – 7 = 0 3x2 – 7 = 0 solve for x2 3x2 = 7 x2 = take square root x = Β±οƒ–7/3 ο‚» Β±1.53 7 3 exact answers approx. answers To solve ax2 + bx + c = 0: 1st: Use the square-root method if the x-term is missing. 2nd: Try factoring it into two binomials. 3rd: Use the quadratic formula (QF). In this course, all approx. answers are given with three significant digits, i.e. starting from the first nonzero digit, round off the answer to a three digit number.
  • 56. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0
  • 57. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0
  • 58. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0 2x(x – 4) = 0
  • 59. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4
  • 60. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0
  • 61. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0
  • 62. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor
  • 63. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3, x = 2 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor and extract the solutions
  • 64. Example G. Solve the by the factoring method. 2. Solve by the factoring method Factor the equation into the form (#x + #)(#x + #) = 0 and obtain a solution from each binomial factor. Solving 2nd Degree Equations ax2 + bx + c = 0 b. x(3x – 3) = –x + 8 3x2 – 3x + x – 8 = 0 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 x = –4/3, x = 2 a. 2x2 – 8x = 0 2x(x – 4) = 0 hence x = 0 or x – 4 = 0 so x = 4 set one side to be 0 factor and extract the solutions If the equation is not easily factorable, use the formula below.
  • 65. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x =
  • 66. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are.
  • 67. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots,
  • 68. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 69. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 70. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 71. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 72. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 73. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 74. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 75. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 76. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 77. –bΒ±οƒ– b2 – 4ac 2a Solving 2nd Degree Equations ax2 + bx + c = 0 Example H. Solve 3x2 – 2x – 2 = 0 Check if it is factorable: a = 3, b = –2, c = –2 So b2 – 4ac = (–2)2 – 4(3)(–2) = 28. Not factorable! Use QF, we get x = – (–2) Β± οƒ– 28 2(3) = 2 Β± οƒ– 28 6 = 2 Β± 2οƒ–7 6 = 1 Β± οƒ–7 3 ο‚» {1.22 –0.549 3. Use the quadratic formula (QF) The roots for the equation ax2 + bx + c = 0 are x = b2 – 4ac is called the discriminant because its value indicates what type of roots there are. Specifically, if b2 – 4ac is a perfect square, we have fractional roots, if b2 – 4ac < 0 there is no real roots.
  • 78. Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0
  • 79. Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20
  • 80. Solving 2nd Degree Equations ax2 + bx + c = 0 Example I. Solve 3x2 – 2x + 2 = 0 Check if it is factorable: a = 3, b = –2, c = 2 So b2 – 4ac = (–2)2 – 4(3)(2) = –20 and –20 does not exist. Hence it does not have any real number solution.
  • 81. Rules of Radicals Exercise 3. A. Solve for x. Give both the exact and approximate answers. If the answer does not exist, state so. 1. x2 = 1 2. x2 – 5 = 4 3. x2 + 5 = 4 4. 2x2 = 31 5. 4x2 – 5 = 4 6. 5 = 3x2 + 1 7. 4x2 = 1 8. x2 – 32 = 42 9. x2 + 62 = 102 10. 2x2 + 7 = 11 11. 2x2 – 5 = 6 12. 4 = 3x2 + 5 B. Solve the following equations by factoring. 5. x2 – 3x = 10 9. x3 – 2x2 = 0 6. x2 = 4 7. 2x(x – 3) + 4 = 2x – 4 10. 2x2(x – 3) = –4x 8. x(x – 3) + x + 6 = 2x2 + 3x 1. x2 – 3x – 4 = 0 2. x2 – 2x – 15 = 0 3. x2 + 7x + 12 = 0 4. –x2 – 2x + 8 = 0 11. 4x2 = x4 12. 7x2 = –4x3 – 3x 13. 5 = (x + 2)(2x + 1) 14. (x + 1)2 = x2 + (x – 1)2 15. (x + 3)2 – (x + 2)2 = (x + 1)2
  • 82. C. Solve the following equations by the quadratic formula. If the answers are not real numbers, just state so. 1. x2 – x + 1 = 0 2. x2 – x – 1 = 0 3. x2 – 3x – 2 = 0 4. x2 – 2x + 3 = 0 5. 2x2 – 3x – 1 = 0 6. 3x2 = 2x + 3 Equations
  • 83. (Answers to odd problems) Exercise C. 1. No real solution 3. π‘₯ = 3 2 Β± 17 2 5. π‘₯ = 1 4 3 Β± 17 1. π‘₯ = βˆ’1, π‘₯ = 4 3. π‘₯ = βˆ’4, π‘₯ = βˆ’3 5. π‘₯ = βˆ’2, π‘₯ = 5 7. π‘₯ = 2 9. π‘₯ = 0, π‘₯ = 2 11. π‘₯ = Β±2, π‘₯ = 0 13. π‘₯ = βˆ’3, π‘₯ = 1 2 15. π‘₯ = 2(2 Β± 5) Equations 1. x = Β±1 3. Doesn’t exist. 7. x = Β±1/2 9. x = Β±8 11. x = ±√11/2 Exercise A. 5. x = Β±3/2 Exercise B.