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g_9 - L_1 Solving Quadratic Equations.pptx
- 1.
- 2. • Solving QuadraticEquations by Extracting the Square Root • Solving Quadratic Equations by Factoring • Solving Quadratic Equations by Completing the Squares • Solving Quadratic Equations by Quadratic Formula
- 3. • illustrates quadraticequations • solves quadratic equations by: (a) extracting square roots; (b) factoring; (c) completing the square; and (d) using the quadratic formula
- 4. A number is8 more than another number. The product of the two numbers is 20. What are the numbers? Second-degree equation (quadratic equations) Let 𝑥 - be the first number 𝑥 + 8 - be the second number 𝑥(𝑥 + 8) = 20 𝑥2 + 8𝑥 = 20
- 5. A quadratic equationsin one variable takes the general form: 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 𝒄 is the constant term 𝒃𝒙 is the linear term 𝒂𝒙𝟐 is the quadratic term Note that: 𝒂 , 𝒃 and 𝒄 are real numbers where 𝑎≠0.
- 6. A quadratic equationcan take any of these forms: 𝒂𝒙𝟐 = 𝒄 𝒂𝒙𝟐 + 𝒃𝒙 = 𝒄 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎 Quadratic equation with no linear term form Standard form General form The values 𝑥 of that make 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 true are referred to as the solutions or roots.
- 7. Solving Quadratic Equationsby Extracting the Square Root The general form of a quadratic equation, 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 = 𝟎, where is 𝒃 = 𝟎 and 𝒂 = 𝟏 can take the form 𝒙𝟐 = 𝟎 and has two possible solutions: 𝒙 = 𝒄 𝒙 = − 𝒄
- 8. Solving Quadratic Equationsby Extracting the Square Root Example 1: 𝒙𝟐 = ± 𝟒𝟗 Solve 𝒙𝟐 = 𝟒𝟗. 𝒙 = ±𝟕 Therefore, the roots of the Quadratic Equation 𝒙𝟐 = 𝟒𝟗 are 7 and -7 Check: 𝒙𝟐 = 𝟒𝟗 (−𝟕)𝟐 = 𝟒𝟗 𝟒𝟗 = 𝟒𝟗 (𝟕)𝟐 = 𝟒𝟗 𝟒𝟗 = 𝟒𝟗 𝒙𝟐 = 𝟒𝟗 𝒙 = 𝟕 𝒙 = −𝟕
- 9. Solving Quadratic Equationsby Extracting the Square Root Example 2: (𝒙 − 𝟓)𝟐= ± 𝟏𝟔 Solve 𝟐(𝒙 − 𝟓)𝟐 = 𝟑𝟐. 𝒙 = 𝟓 ± 𝟒 𝟐(𝒙 − 𝟓)𝟐 𝟐 = 𝟑𝟐 𝟐 (𝒙 − 𝟓)𝟐 = 𝟏𝟔 𝒙 = 𝟓 + 𝟒 𝒙 = 𝟓 − 𝟒 𝒙 = 𝟗 𝒙 = 𝟏 𝒙 − 𝟓 = ±𝟒
- 10. Solving Quadratic Equationsby Extracting the Square Root Example 2: Solve 𝟐(𝒙 − 𝟓)𝟐 = 𝟑𝟐. Therefore, the roots of the Quadratic Equation 𝟐(𝒙 − 𝟓)𝟐 = 𝟑𝟐 are 9 and 1 Check: 𝟐(𝒙 − 𝟓)𝟐 = 𝟑𝟐 𝟐(𝟗 − 𝟓)𝟐 = 𝟑𝟐 𝟐(𝟒)𝟐 = 𝟑𝟐 𝟐(𝟏𝟔) = 𝟑𝟐 𝟑𝟐 = 𝟑𝟐 𝟐(𝒙 − 𝟓)𝟐 = 𝟑𝟐 𝟐(𝟏 − 𝟓)𝟐 = 𝟑𝟐 𝟐(−𝟒)𝟐 = 𝟑𝟐 𝟐(𝟏𝟔) = 𝟑𝟐 𝟑𝟐 = 𝟑𝟐
- 11. In solving quadraticequations by factoring, we apply the principle of zero product to determine the solutions. In the principle of zero product, if𝒂𝒃 = 𝟎, then 𝒂 = 𝟎 or 𝒃 = 𝟎.
- 12. Example 1: Solve thequadratic equation 𝟗𝒙𝟐 = 𝟐𝟓 by factoring. Solution: 𝟗𝒙𝟐 = 𝟐𝟓 𝟗𝒙𝟐 − 𝟐𝟓 = 𝟎 (𝟑𝒙 + 𝟓)(𝟑𝒙 − 𝟓) = 𝟎 (𝟑𝒙 + 𝟓) = 𝟎 (𝟑𝒙 − 𝟓) = 𝟎 𝟑𝒙 = −𝟓 𝟑𝒙 𝟑 = −𝟓 𝟑 𝟑𝒙 = 𝟓 𝒙 = −𝟓 𝟑 𝟑𝒙 𝟑 = 𝟓 𝟑 𝒙 = 𝟓 𝟑
- 13. Example 1: Solve thequadratic equation 𝟗𝒙𝟐 = 𝟐𝟓 by factoring. Check: 𝟗𝒙𝟐 = 𝟐𝟓 𝟗(−𝟓 𝟑 )𝟐 = 𝟐𝟓 𝟗(𝟐𝟓 𝟗 ) = 𝟐𝟓 𝟐𝟓 = 𝟐𝟓 𝟗𝒙𝟐 = 𝟐𝟓 𝟗(𝟓 𝟑 )𝟐 = 𝟐𝟓 𝟗(𝟐𝟓 𝟗 ) = 𝟐𝟓 𝟐𝟓 = 𝟐𝟓 Therefore, the roots of the Quadratic Equation 𝟗𝒙𝟐 = 𝟐𝟓 are −𝟓 𝟑 and 𝟓 𝟑
- 14. Example 2: Solve thequadratic equation 𝟒𝒙𝟐 + 𝟕𝒙 = 𝟎 . Solution: 𝟒𝒙𝟐 + 𝟕𝒙 = 𝟎 𝒙(𝟒𝒙 + 𝟕) = 𝟎 𝒙 = 𝟎 𝟒𝒙 + 𝟕 = 𝟎 𝟒𝒙 = −𝟕 𝟒𝒙 𝟒 = − 𝟕 𝟒 𝒙 = −𝟕 𝟒 𝟒𝒙 + 𝟕 = 𝟎 −𝟕 + 𝟕 = 𝟎 𝟎 = 𝟎 𝟒(−𝟕 𝟒 ) + 𝟕 = 𝟎 Therefore, the roots of the Quadratic Equation 𝟒𝒙𝟐 + 𝟕𝒙 = 𝟎 are 𝟎 and −𝟕 𝟒
- 15. Example 3: Solve theroots of the equation 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟑 = 𝟎 . Solution: 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟑 = 𝟎 (𝟐𝒙 + 𝟏)(𝒙 − 𝟑) = 𝟎 𝟐𝒙 + 𝟏 = 𝟎 𝟐𝒙 = −𝟏 𝟐𝒙 𝟐 = −𝟏 𝟐 𝒙 = −𝟏 𝟐 Therefore, the roots of the Quadratic Equation 𝟐𝒙𝟐 − 𝟓𝒙 − 𝟑 = 𝟎 are −𝟏 𝟐 and 𝟑 𝒙 − 𝟑 = 𝟎 𝒙 = 𝟑
- 16. In solving quadraticequations by completing the square, we make the left- hand side of the equation a complete perfect square trinomial by applying the addition property of equality. 𝒙𝟐 + 𝟔𝒙 = 𝟖 Step 1: Divide the numerical coefficient of the linear term by 2 Step 2: Square the quotient in Step 1 Step 3: Then, add the result in Step 2 to both sides of the equation 6 2 = 𝟑 32 = 𝟗 𝑥2 + 𝟔𝒙 + 𝟗 = 𝟖 + 𝟗 𝑥2 + 𝟔𝒙 + 𝟗 = 𝟏𝟕
- 17. Example: 𝟐𝒙𝟐 − 𝟑𝒙− 𝟗 = 𝟎 𝟐𝒙𝟐 − 𝟑𝒙 = 𝟗 𝟐 𝟐𝒙𝟐 − 𝟑 𝟐𝒙 = 𝟗 𝟐 𝒙𝟐 − 𝟑 𝟐𝒙 + 𝟗 𝟏𝟔 = 𝟗 𝟐+ 𝟗 𝟏𝟔 (𝒙 − 𝟑 𝟒)𝟐= 𝟕𝟐+𝟗 𝟏𝟔 (𝒙 − 𝟑 𝟒)𝟐 = 𝟖𝟏 𝟏𝟔 (𝒙 − 𝟑 𝟒)𝟐= ± 𝟖𝟏 𝟏𝟔 𝒙 − 𝟑 𝟒 = ± 𝟗 𝟒 𝒙 = 𝟑 𝟒 ± 𝟗 𝟒 Thus, the square roots of the equation are 𝟑 or − 𝟑 𝟐 𝒙 = 𝟑 𝟒 − 𝟗 𝟒 𝒙 = 𝟑 𝟒 + 𝟗 𝟒
- 18. Another way ofsolving quadratic equations is using the quadratic formula: 𝒙 = −𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄 𝟐𝒂
- 19. Example: 𝒙𝟐 − 𝟕𝒙= −𝟏𝟎 𝒙𝟐 − 𝟕𝒙 + 𝟏𝟎 = 𝟎 𝒂 = 𝟏, 𝒃 = −𝟕, 𝒄 = 𝟏𝟎 𝒙 = −𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄 𝟐𝒂 𝒙 = −(−𝟕) ± (−𝟕)𝟐−𝟒(𝟏)(𝟏𝟎) 𝟐(𝟏) 𝒙 = 𝟕 ± 𝟒𝟗 − 𝟒𝟎 𝟐 𝒙 = 𝟕 ± 𝟗 𝟐 𝒙 = 𝟕 ± 𝟑 𝟐 𝒙 = 𝟕 + 𝟑 𝟐 𝒙 = 𝟕 − 𝟑 𝟐 𝒙 = 𝟐 𝒙 = 𝟓
- 20. Check: 𝒙 = 𝟕 +𝟑 𝟐 𝒙 = 𝟕 − 𝟑 𝟐 𝒙 = 𝟐 𝒙 = 𝟓 𝒙𝟐 − 𝟕𝒙 = −𝟏𝟎 𝒙𝟐 − 𝟕𝒙 = −𝟏𝟎 𝟓𝟐 − 𝟕(𝟓) = −𝟏𝟎 𝟐𝟓 − 𝟑𝟓 = −𝟏𝟎 −𝟏𝟎 = −𝟏𝟎 𝟐𝟐 − 𝟕(𝟐) = −𝟏𝟎 𝟒 − 𝟏𝟒 = −𝟏𝟎 −𝟏𝟎 = −𝟏𝟎 Thus, the square roots of the equation 𝒙𝟐 − 𝟕𝒙 = −𝟏𝟎 are 𝟓 and 𝟐
- 21. Seatwork No. 1 Solveby Extracting the Square Root. Solve by Factoring. Solve by Completing the Square. Solve by using Quadratic Formula. 𝒙𝟐 = 𝟔𝟒 𝟑𝒙𝟐 = 𝟏𝟎𝟖 𝒙𝟐 − 𝟔 = 𝟎 𝒙𝟐 − 𝟐𝟏 = 𝟎 𝒎𝟐 + 𝟏𝟔𝒎 = 𝟒 𝒙𝟐 − 𝟔 = 𝟖 𝒙𝟐 + 𝟏𝟎 + 𝟗 = 𝟎 𝒙𝟐 + 𝟕 + 𝟏𝟎 = 𝟎