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Quadratic Equations: Square Roots and Completing the Square

Learn how to solve quadratic equations by finding square roots and completing the square. Practice problems included.

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Quadratic Equations: Square Roots and Completing the Square

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  1. Chapter 4 Section 7

  2. ANSWER The solutions are 4 + 5 = 9 and 4 –5 = – 1. EXAMPLE 1 Solve a quadratic equation by finding square roots Solve x2 – 8x + 16 = 25. x2 – 8x + 16 = 25 Write original equation. (x – 4)2 = 25 Write left side as a binomial squared. x – 4 = +5 Take square roots of each side. x = 4 + 5 Solve for x.

  3. 16 = 8 2 EXAMPLE 2 Make a perfect square trinomial Find the value of cthat makes x2 + 16x + ca perfect square trinomial. Then write the expression as the square of a binomial. SOLUTION STEP 1 Find half the coefficient of x. STEP 2 Square the result of Step 1. 82 = 64 STEP 3 Replace cwith the result of Step 2. x2 + 16x + 64

  4. EXAMPLE 2 Make a perfect square trinomial ANSWER The trinomialx2 + 16x + c is a perfect square whenc = 64. Thenx2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

  5. for Examples 1 and 2 GUIDED PRACTICE Solve the equation by finding square roots. 1. x2 + 6x + 9 = 36. ANSWER 3 and –9. 2. x2 – 10x + 25 = 1. ANSWER 4 and 6. 3. x2 – 24x + 144 = 100. ANSWER 2 and 22.

  6. 81 9 4 2 for Examples 1 and 2 GUIDED PRACTICE Find the value of c that makes the expression a perfect square trinomial.Then write the expression as the square of a binomial. x2 + 14x + c 4. ANSWER 49 ; (x + 7)2 x2 + 22x + c 5. ANSWER 121 ; (x + 11)2 x2 – 9x + c 6. ANSWER ; (x – )2.

  7. ) ( –12 Add to each side. 2 36 (–6) 2 = = 2 x – 6 = + 32 x = 6 + 32 x = 6 + 4 2 2 Simplify: 32 16 2 4 = = ANSWER 2 2 The solutions are 6 + 4 and6 – 4 EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1 Solve x2 – 12x + 4 = 0 by completing the square. x2 – 12x + 4 = 0 Write original equation. x2 – 12x = –4 Write left side in the form x2 + bx. x2– 12x + 36 = –4 + 36 (x – 6)2 = 32 Write left side as a binomial squared. Take square roots of each side. Solve for x.

  8. You can use algebra or a graph. Algebra Substitute each solution in the original equation to verify that it is correct. Graph Use a graphing calculator to graph y = x2 – 12x + 4. The x-intercepts are about 0.34 6 – 4 2 and 11.66 6 + 4 2 EXAMPLE 3 Solve ax2 + bx + c = 0 when a = 1 CHECK

  9. Solve ax2 + bx + c = 0 when a = 1 ) ( 4 Add to each side. 2 4 2 2 = = 2 x + 2 = + –3 x = –2 + –3 x = –2 +i 3 EXAMPLE 4 Solve 2x2 + 8x + 14 = 0 by completing the square. 2x2 + 8x + 14 = 0 Write original equation. x2 + 4x + 7 = 0 Divide each side by the coefficient of x2. x2 + 4x = –7 Write left side in the form x2 + bx. x2– 4x + 4 = –7 + 4 (x + 2)2 = –3 Write left side as a binomial squared. Take square roots of each side. Solve for x. Write in terms of the imaginary unit i.

  10. Solve ax2 + bx + c = 0 when a = 1 ANSWER The solutions are –2 + i and–2 – i 3 3 . EXAMPLE 4

  11. EXAMPLE 5 Standardized Test Practice SOLUTION Use the formula for the area of a rectangle to write an equation.

  12. Length Width = Area ) ( 2 Add to each side. 2 1 1 2 = = 2 EXAMPLE 5 Standardized Test Practice 3x(x + 2) = 72 3x2 + 6x = 72 Distributive property x2 + 2x = 24 Divide each side by the coefficient of x2. x2– 2x + 1 = 24 + 1 (x + 1)2 = 25 Write left side as a binomial squared. x + 1 = + 5 Take square roots of each side. x = –1 + 5 Solve for x.

  13. ANSWER The value of xis 4. The correct answer is B. EXAMPLE 5 Standardized Test Practice So, x = –1 + 5 = 4 or x = – 1 – 5 = –6. You can rejectx = –6because the side lengths would be–18and–4, and side lengths cannot be negative.

  14. ANSWER –3+ 5 ANSWER –2 + 10 ANSWER 5 + 17 ANSWER –4 +3 2 ANSWER ANSWER 1 + 2 2 1 + 26 for Examples 3, 4 and 5 GUIDED PRACTICE Solve the equation by completing the square. 7. 10. 3x2 + 12x – 18 = 0 x2 + 6x + 4 = 0 8. x2 – 10x + 8 = 0 11. 6x(x + 8) = 12 9. 12. 4p(p – 2) = 100 2n2 – 4n – 14 = 0

  15. ) ( –10 Add to each side. 2 25 (–5) 2 = = 2 ANSWER The vertex form of the function is y = (x – 5)2– 3. The vertex is (5, –3). EXAMPLE 6 Write a quadratic function in vertex form Write y = x2 – 10x + 22 in vertex form. Then identify the vertex. y = x2 – 10x + 22 Write original function. y + ?= (x2–10x + ?) + 22 Prepare to complete the square. y + 25= (x2– 10x + 25) + 22 y + 25 = (x – 5)2 + 22 Write x2 – 10x + 25 as a binomial squared. y = (x – 5)2– 3 Solve for y.

  16. EXAMPLE 7 Find the maximum value of a quadratic function Baseball The height y(in feet) of a baseball tseconds after it is hit is given by this function: y = –16t2 + 96t + 3 Find the maximum height of the baseball. SOLUTION The maximum height of the baseball is the y-coordinate of the vertex of the parabola with the given equation.

  17. (–16)(9) Add to each side. ANSWER The vertex is (3, 147), so the maximum height of the baseball is 147feet. EXAMPLE 7 Find the maximum value of a quadratic function y = –16t2+ 96t +3 Write original function. y = –16(t2– 6t)+3 Factor –16 from first two terms. y +(–16)(?)= –16(t2 –6t + ?) + 3 Prepare to complete the square. y +(–16)(9)= –16(t2 – 6t + 9) + 3 y – 144 = –16(t – 3)2 + 3 Write t2 – 6t + 9 as a binomial squared. y = –16(t – 3)2+ 147 Solve for y.

  18. for Examples 6 and 7 GUIDED PRACTICE Write the quadratic function in vertex form. Then identify the vertex. 13. y = x2 – 8x + 17 y = (x – 4)2+ 1 ; (4, 1). ANSWER 14. y = x2 + 6x + 3 y = (x + 3)2– 6 ; (–3, –6) ANSWER 15. f(x) = x2 – 4x – 4 ANSWER y = (x – 2)2– 8 ; (2 , –8)

  19. for Examples 6 and 7 GUIDED PRACTICE 16. What if ? In example 7, suppose the height of the baseball is given by y = – 16t2 + 80t + 2. Find the maximum height of the baseball. ANSWER 102feet.

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