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First Quarter - Chapter 2 - Quadratic Equation

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- 1. Quadratic Equation
- 2. *1 – Definition of Quadratic Equation2 – Solving Quadratic Equations by the Square Root Property3 – Solving Quadratic Equations by Completing the Square4 – Solving Quadratic Equations by the Quadratic Formula5 – Graphing Quadratic Equations in Two Variables6 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions
- 3. *Quadratic Equations*An example of a Quadratic Equation:*The name Quadratic comes from "quad" meaning square, because the variable gets squared (like x2).*It is also called an "Equation of Degree 2" (because of the "2" on the x)
- 4. The Standard Form of a Quadratic Equation looks like this: The letters a, b and c are coefficients (you know those values). They can have any value, except that a cant be 0. The letter "x" is the variable or unknown (you dont know it yet)
- 5. Solving Quadratic Equationsby the Square Root Property
- 6. *We previously have used factoring to solve quadraticequations.This chapter will introduce additional methods for solvingquadratic equations.Square Root Property If b is a real number and a2 = b, then a b
- 7. Example Solve x2 = 49 x 49 7 Solve 2x2 = 4 x2 = 2 x 2 Solve (y – 3)2 = 4 y 3 4 2 y=3 2 y = 1 or 5
- 8. Example Solve x2 + 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number.
- 9. Example Solve (x + 2)2 = 25 x 2 25 5 x= 2 5 x = 2 + 5 or x = 2 – 5 x = 3 or x = 7
- 10. Example Solve (3x – 17)2 = 28 3x – 17 = 28 2 7 3x 17 2 7 17 2 7 x 3
- 11. Solving QuadraticEquations by Completingthe Square
- 12. Completing the SquareIn all four of the previous examples, theconstant in the square on the right side, is halfthe coefficient of the x term on the left.Also, the constant on the left is the square ofthe constant on the right.So, to find the constant term of a perfect squaretrinomial, we need to take the square of half thecoefficient of the x term in the trinomial (as longas the coefficient of the x2 term is 1, as in ourprevious examples).
- 13. Example What constant term should be added to the following expressions to create a perfect square trinomial? x – 10x 2 add 52 = 25 x2 + 16x add 82 = 64 x2 – 7x 2 7 49 add 2 4
- 14. Example We now look at a method for solving quadratics that involves a technique called completing the square. It involves creating a trinomial that is a perfect square, setting the factored trinomial equal to a constant, then using the square root property from the previous section.
- 15. Solving a Quadratic Equation by Completing aSquare 1) If the coefficient of x2 is NOT 1, divide both sides of the equation by the coefficient. 2) Isolate all variable terms on one side of the equation. 3) Complete the square (half the coefficient of the x term squared, added to both sides of the equation). 4) Factor the resulting trinomial. 5) Use the square root property.
- 16. Solving EquationsExample Solve by completing the square. y2 + 6y = 8 y2 + 6y + 9 = 8 + 9 (y + 3)2 = 1 y+3= 1 = 1 y= 3 1 y = 4 or 2
- 17. Example Solve by completing the square. y2 + y – 7 = 0 y2 + y = 7 y2 + y + ¼ = 7 + ¼ 29 (y + ½)2 = 4 1 29 29 y 2 4 2 1 29 1 29 y 2 2 2
- 18. ExampleSolve by completing the square. 2x2 + 14x – 1 = 0 2x2 + 14x = 1 x2 + 7x = ½ 49 49 51 x2 + 7x + 4 =½+ 4 = 4 7 51 (x + )2 = 2 4 7 51 51 7 51 7 51 x x 2 4 2 2 2 2
- 19. Solving QuadraticEquations by theQuadratic Formula
- 20. *Another technique for solving quadratic equations is to use the quadratic formula.The formula is derived from completing the square of a general quadratic equation.
- 21. A quadratic equation written in standard form, ax2 + bx + c = 0, has the solutions. 2 b b 4 ac x 2a
- 22. Example Solve 11n2 – 9n = 1 by the quadratic formula. 11n2 – 9n – 1 = 0, so a = 11, b = -9, c = -1 2 9 ( 9) 4 (11 )( 1) n 2 (11 ) 9 81 44 9 125 9 5 5 22 22 22
- 23. Example 1 5Solve x2 + x – = 0 by the 8 2quadratic formula. x2 + 8x – 20 = 0 (multiply both sides by 8) a = 1, b = 8, c = 20 2 8 (8 ) 4 (1)( 20 ) 8 64 80 8 144x 2 (1) 2 2 8 12 20 4 or , 10 or 2 2 2 2
- 24. Example Solve x(x + 6) = 30 by the quadratic formula. x2 + 6x + 30 = 0 a = 1, b = 6, c = 30 2 6 (6) 4 (1)( 30 ) 6 36 120 6 84x 2 (1) 2 2 So there is no real solution.
- 25. *The expression under the radical sign in theformula (b2 – 4ac) is called the discriminant.The discriminant will take on a value that ispositive, 0, or negative.The value of the discriminant indicates twodistinct real solutions, one real solution, or noreal solutions, respectively.
- 26. ExampleUse the discriminant to determine the number and type ofsolutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224There are no real solutions.
- 27. Example 2: Using the Discriminant Find the number of solutions of each equation using the discriminant.A. B. C.3x2 – 2x + 2 = 0 2x2 + 11x + 12 = 0 x2 + 8x + 16 = 0a = 3, b = –2, c = 2 a = 2, b = 11, c = 12 a = 1, b = 8, c = 16 b2 – 4ac b2 – 4ac b2 – 4ac(–2)2 – 4(3)(2) 112 – 4(2)(12) 82 – 4(1)(16) 4 – 24 121 – 96 64 – 64 –20 25 0b2 – 4ac is negative. b2 – 4ac is positive. b2 – 4ac is zero. There are no real There are two real There is one real solution solutions solutions
- 28. Check It Out! Example 2 Find the number of solutions of each equation using the discdriminant.a. b. c.2x2 – 2x + 3 = 0 x2 + 4x + 4 = 0 x2 – 9x + 4 = 0a = 2, b = –2, c = 3 a = 1, b = 4, c = 4 a = 1, b = –9 , c = 4 b2 – 4ac b2 – 4ac b2 – 4ac (–2)2 – 4(2)(3) 42 – 4(1)(4) (–9)2 – 4(1)(4) 4 – 24 16 – 16 81 – 16 –20 0 65b2 – 4ac is negative. b2 – 4ac is zero. b2 – 4ac is positive. There are no real There is one real solution There are two real solutions solutions
- 29. ApplicationThe height h in feet of an object shot straight up withinitial velocity v in feet per second is given by h = –16t2+ vt + c, where c is the initial height of the objectabove the ground. The ringer on a carnival strength testis 2 feet off the ground and is shot upward with aninitial velocity of 30 feet per second. Will it reach aheight of 20 feet? Use the discriminant to explain youranswer.
- 30. Continued h = –16t2 + vt + c Substitute 20 for h, 30 for v, and 2 for c.20 = –16t2 + 30t + 2 0 = –16t2 + 30t + (–18) Subtract 20 from both sides. b2 – 4ac Evaluate the discriminant.302 – 4(–16)(–18) = –252 Substitute –16 for a, 30 for b, and –18 for c. The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.
- 31. Check It Out! Example 2 What if…? Suppose the weight is shot straight up with an initial velocity of 20 feet per second from 1 foot above the ground. Will it ring the bell? Use the discriminant to explain your answer. h = –16t2 + vt + c20 = –16t2 + 20t + 1 Substitute 20 for h, 20 for v, and 1 for c. 0 = –16t2 + 20t + (–19) Subtract 20 from both sides. b2 – 4ac Evaluate the discriminant.202 – 4(–16)(–19) = –816 Substitute –16 for a, 20 for b, and –19 for c. The discriminant is negative, so there are no real solutions. The ringer will not reach a height of 20 feet.

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