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Section 5.7 Completing the Square

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**Simplify each expression.**

Completing the Square ALGEBRA 2 LESSON 5-7 (For help, go to Lessons 5-1 and page 262.) 9 16 1. (x – 3)(x – 3) 2. (2x – 1)(2x – 1) 3. (x + 4)(x + 4) – 3 4. ± 25 5. ± ± –4 7. ± Simplify each expression. Check Skills You’ll Need 5-7

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**1. (x – 3)(x – 3) = x2 – 2(3)x + 32 = x2 – 6x + 9 **

Completing the Square ALGEBRA 2 LESSON 5-7 Solutions 9 16 3 4 1. (x – 3)(x – 3) = x2 – 2(3)x + 32 = x2 – 6x + 9 2. (2x – 1)(2x – 1) = (2x)2 – 2(1)(2x) + 12 = 4x2 – 4x + 1 3. (x + 4)(x + 4) – 3 = x2 + 2(4)x + 42 – 3 = x2 + 8x + 16 – 3 = x2 + 8x + 13 4. ± = ±5 5. ± 48 = ± • 3 = ±4 3 6. ± –4 = ± –1 • 4 = ±2i 7. ± = ± = ± 5-7

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**(x – 6)2 = 9 Factor the trinomial.**

Completing the Square ALGEBRA 2 LESSON 5-7 Solve x2 – 12x + 36 = 9. x2 – 12x + 36 = 9 (x – 6)2 = 9 Factor the trinomial. x – 6 = ±3 Find the square root of each side. x – 6 = 3 or x – 6 = –3 Solve for x. x = 9 or x = 3 Quick Check 5-7

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**Find the missing value to complete the square: x2 + 20x + .**

Completing the Square ALGEBRA 2 LESSON 5-7 Find the missing value to complete the square: x2 + 20x = = 100 Find Substitute 20 for b. b 2 20 x2 + 20x Complete the square. Quick Check 5-7

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**x2 + 4x = –1 Rewrite so all terms containing x are on one side.**

Completing the Square ALGEBRA 2 LESSON 5-7 Solve x2 + 4x + 1 = 0. 4 2 = 4 Find 2 b 2 2 x2 + 4x = –1 Rewrite so all terms containing x are on one side. x2 + 4x + 4 = –1 + 4 Complete the square by adding 4 to each side. (x + 2)2 = 3 Factor the perfect square trinomal. x + 2 = ± 3 Find the square root of each side. x = –2 ± 3 Solve for x. 5-7

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**Completing the Square (continued) Check:**

ALGEBRA 2 LESSON 5-7 (continued) Check: (–2)2 – 2(–2 3) + ( 3)2 – 8 – (– )2 + 4(– ) + 1 x2 + 4x + 1 0 = 0 4 – – (4 + 3 – 8 + 1) + (– ) (–2)2 + 2(–2 3) + ( 3)2 + (–8) (–2 – 3)2 + 4(–2 – ) + 1 – 8 – (4 + 3 – 8 + 1) + ( – ) Quick Check 5-7

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**Rewrite so all terms containing x are on one side.**

Completing the Square ALGEBRA 2 LESSON 5-7 Solve x2 + 6x + 12 = 0. 6 2 2 b 2 2 = 9 Find x2 + 6x = –12 Rewrite so all terms containing x are on one side. x2 + 6x + 9 = –12 + 9 Complete the square by adding 9 to each side. (x + 3)2 = –3 Factor the perfect square trinomial. x + 3 = ± –3 Find the square root of each side. x = –3 ± –3 Solve for x. = –3 ± i 3 Simplify. Quick Check 5-7

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**= Completing the Square Solve 2x2 + 7x – 1 = 0. x2 + x – = 0**

ALGEBRA 2 LESSON 5-7 Solve 2x2 + 7x – 1 = 0. x x – = 0 7 2 1 Divide each side by 2. x x = 1 2 7 Rewrite so all terms containing x are on one side. = 7 2 49 16 Find b x x = + 7 2 49 16 1 Complete the square by adding to each side. x = 7 4 57 16 Factor the perfect square trinomial. 2 7 4 57 x + = ± Find the square root of each side. x = – ± 7 4 Solve for x. 57 Quick Check 5-7

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**Write y = x2 + 5x + 2 in vertex form.**

Completing the Square ALGEBRA 2 LESSON 5-7 Write y = x2 + 5x + 2 in vertex form. y = x2 + 5x + 2 = x2 + 5x – 5 2 Complete the square. Add and subtract on the right side. = x – 25 4 5 2 Factor the perfect square trinomial. = x – 5 2 17 4 Simplify. The vertex form is y = x – . 5 2 17 4 Quick Check 5-7

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**Factor –1 from the first two terms.**

Completing the Square ALGEBRA 2 LESSON 5-7 Quick Check The monthly profit P from the sales of rugs woven by a family rug-making business depends on the price r that they charge for a rug. The profit is model by P = –r r – 59,500. Write the function in vertex form. Use the vertex form to find the price that yields the maximum monthly profit and the amount of the maximum profit. P = –r r – 59500 = –(r 2 – 500r) – 59500 Factor –1 from the first two terms. = –[r 2 – 500r + (–250)2] – (–250)2 Add and subtract (–250)2 on the right side. = –(r – 250)2 – Factor the perfect square trinomial. = –(r – 250) Simplify in vertex form. The vertex is (250, 3000). A price of $250 per rug gives a maximum monthly profit of $3000. 5-7

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**Simplify each expression. 3. x2 – 6x – 16 = 0 4. x2 – 14x + 74 = 0**

Completing the Square ALGEBRA 2 LESSON 5-7 Complete the square. 1. x2 + 60x + 2. x2 – 7x + Simplify each expression. 3. x2 – 6x – 16 = 0 4. x2 – 14x + 74 = 0 5. 3x2 + 5x – 28 = x2 – 6x + 3 = 0 49 9 900 –2, 8 7 ± 5i –4, 7 3 3 4 ± i 5-7

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Why we complete the square We have learned how to factor quadratic expressions to solve. Many quadratic equations contain expressions that cannot be.

Why we complete the square We have learned how to factor quadratic expressions to solve. Many quadratic equations contain expressions that cannot be.

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