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First Quarter - Chapter 2 - Quadratic Equation
Okay, let's think through this with the new information: * The equation modeling the height is: h = -16t^2 + vt + c * The initial height (c) is still 2 feet * The initial velocity (v) is now 20 feet/second * The target height (h) is still 20 feet So the equation is: 20 = -16t^2 + 20t + 2 0 = -16t^2 + 20t + 18 (subtract 20 from both sides) Evaluating the discriminant: (20)^2 - 4(-16)(-18) = 400 - 288 = 112 Since the discriminant is positive
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Similar to First Quarter - Chapter 2 - Quadratic Equation
First Quarter - Chapter 2 - Quadratic Equation
- 1.
- 2. * 1 – Definitionof Quadratic Equation 2 – Solving Quadratic Equations by the Square Root Property 3 – Solving Quadratic Equations by Completing the Square 4 – Solving Quadratic Equations by the Quadratic Formula 5 – Graphing Quadratic Equations in Two Variables 6 – Interval Notation, Finding Domains and Ranges from Graphs and Graphing Piecewise-Defined Functions
- 3. *Quadratic Equations *An exampleof 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 Formof 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 can't be 0. The letter "x" is the variable or unknown (you don't know it yet)
- 7. Solving Quadratic Equations bythe Square Root Property
- 8. * We previously haveused factoring to solve quadratic equations. This chapter will introduce additional methods for solving quadratic equations. Square Root Property If b is a real number and a2 = b, then a b
- 9. 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
- 10. Example Solve x2+ 4 = 0 x2 = 4 There is no real solution because the square root of 4 is not a real number.
- 11. 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
- 12. Example Solve (3x– 17)2 = 28 3x – 17 = 28 2 7 3x 17 2 7 17 2 7 x 3
- 13. Solving Quadratic Equations byCompleting the Square
- 14. Completing the Square Inall four of the previous examples, the constant in the square on the right side, is half the coefficient of the x term on the left. Also, the constant on the left is the square of the constant on the right. So, to find the constant term of a perfect square trinomial, we need to take the square of half the coefficient of the x term in the trinomial (as long as the coefficient of the x2 term is 1, as in our previous examples).
- 15. Example What constantterm 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
- 16. Example We nowlook 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.
- 17. Solving a QuadraticEquation by Completing a Square 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.
- 18. Solving Equations Example 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
- 19. Example Solveby 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
- 20. Example Solve by completingthe 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
- 21. Solving Quadratic Equations bythe Quadratic Formula
- 22. * Another technique forsolving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.
- 23. A quadratic equationwritten in standard form, ax2 + bx + c = 0, has the solutions. 2 b b 4 ac x 2a
- 24. 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
- 25. Example 1 5 Solve x2 + x – = 0 by the 8 2 quadratic 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 8 144 x 2 (1) 2 2 8 12 20 4 or , 10 or 2 2 2 2
- 26. 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 84 x 2 (1) 2 2 So there is no real solution.
- 27. * The expression underthe radical sign in the formula (b2 – 4ac) is called the discriminant. The discriminant will take on a value that is positive, 0, or negative. The value of the discriminant indicates two distinct real solutions, one real solution, or no real solutions, respectively.
- 28. Example Use the discriminantto determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.
- 30. Example 2: Usingthe 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 = 0 a = 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 0 b2 – 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
- 31. 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 = 0 a = 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 65 b2 – 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
- 32. Application The height hin feet of an object shot straight up with initial velocity v in feet per second is given by h = –16t2 + vt + c, where c is the initial height of the object above the ground. The ringer on a carnival strength test is 2 feet off the ground and is shot upward with an initial velocity of 30 feet per second. Will it reach a height of 20 feet? Use the discriminant to explain your answer.
- 33. 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.
- 34. 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 + c 20 = –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.