1. 2.

Objective Solve quadratic equations by completing the square.

  “b divided by two, squared”   X2 + 6x + 9 x2 – 8x + 16 Divide the coefficient of the x-term by 2, then square the result to get the constant term.

An expression in the form x2 + bx is not a perfect square An expression in the form x2 + bx is not a perfect square. However, you can use the relationship shown above to add a term to x2 + bx to form a trinomial that is a perfect square. This is called completing the square.

Example 1: Completing the Square Complete the square to form a perfect square trinomial. A. x2 + 2x + B. x2 – 6x + x2 + 2x x2 + –6x Identify b. . x2 + 2x + 1 x2 – 6x + 9

Check It Out! Example 1 Complete the square to form a perfect square trinomial. c. x2 + 12x + d. 8x + x2 + x2 + 12x x2 + 8x Identify b. . x2 + 12x + 36 x2 + 12x + 16

 

Solving a Quadratic Equation by Completing the Square

Example 2A: Solving x2 +bx = c Solve by completing the square. x2 + 16x = –15 The equation is in the form x2 + bx = c. Step 1 x2 + 16x = –15 Step 2 . Step 3 x2 + 16x + 64 = –15 + 64 Complete the square. Step 4 (x + 8)2 = 49 Factor and simplify. Take the square root of both sides. Step 5 x + 8 = ± 7 Step 6 x + 8 = 7 or x + 8 = –7 x = –1 or x = –15 Write and solve two equations.

  Example 2A Continued Solve by completing the square. x2 + 16x = –15 The solutions are –1 and –15. Check x2 + 16x = –15 (–1)2 + 16(–1) –15 1 – 16 –15 –15 –15  x2 + 16x = –15 (–15)2 + 16(–15) –15 225 – 240 –15 –15 –15 

Example 2B: Solving x2 +bx = c Solve by completing the square. x2 – 4x – 6 = 0 Write in the form x2 + bx = c. Step 1 x2 + (–4x) = 6 Step 2 . Step 3 x2 – 4x + 4 = 6 + 4 Complete the square. Step 4 (x – 2)2 = 10 Factor and simplify. Take the square root of both sides. Step 5 x – 2 = ± √10 Step 6 x – 2 = √10 or x – 2 = –√10 x = 2 + √10 or x = 2 – √10 Write and solve two equations.

Example 2B Continued Solve by completing the square. The solutions are 2 + √10 and x = 2 – √10. Check Use a graphing calculator to check your answer.

Check It Out! Example 2a Solve by completing the square. x2 + 10x = –9 The equation is in the form x2 + bx = c. Step 1 x2 + 10x = –9 Step 2 . Step 3 x2 + 10x + 25 = –9 + 25 Complete the square. Factor and simplify. Step 4 (x + 5)2 = 16 Take the square root of both sides. Step 5 x + 5 = ± 4 Step 6 x + 5 = 4 or x + 5 = –4 x = –1 or x = –9 Write and solve two equations.

Check It Out! Example 2a Continued Solve by completing the square. x2 + 10x = –9 The solutions are –9 and –1. Check x2 + 16x = –15 (–1)2 + 16(–1) –15 1 – 16 –15 –15 –15  x2 + 10x = –9 (–9)2 + 10(–9) –9 81 – 90 –9 –9 –9 

Example 3A: Solving ax2 + bx = c by Completing the Square Solve by completing the square. –3x2 + 12x – 15 = 0 Step 1 Divide by – 3 to make a = 1. x2 – 4x + 5 = 0 x2 – 4x = –5 Write in the form x2 + bx = c. x2 + (–4x) = –5 Step 2 . Step 3 x2 – 4x + 4 = –5 + 4 Complete the square.

Example 3A Continued Solve by completing the square. –3x2 + 12x – 15 = 0 Step 4 (x – 2)2 = –1 Factor and simplify. There is no real number whose square is negative, so there are no real solutions.

Understand the Problem Example 4: Problem-Solving Application A rectangular room has an area of 195 square feet. Its width is 2 feet shorter than its length. Find the dimensions of the room. Round to the nearest hundredth of a foot, if necessary. 1 Understand the Problem The answer will be the length and width of the room. List the important information: The room area is 195 square feet. The width is 2 feet less than the length.

Example 4 Continued 2 Make a Plan Set the formula for the area of a rectangle equal to 195, the area of the room. Solve the equation.

Use the formula for area of a rectangle. Example 4 Continued Solve 3 Let x be the width. Then x + 2 is the length. Use the formula for area of a rectangle. l • w = A length times width = area of room x + 2 • x = 195

Example 4 Continued Step 1 x2 + 2x = 195 Simplify. . Step 2 Complete the square by adding 1 to both sides. Step 3 x2 + 2x + 1 = 195 + 1 Step 4 (x + 1)2 = 196 Factor the perfect-square trinomial. Take the square root of both sides. Step 5 x + 1 = ± 14 Step 6 x + 1 = 14 or x + 1 = –14 Write and solve two equations. x = 13 or x = –15

Example 4 Continued Negative numbers are not reasonable for length, so x = 13 is the only solution that makes sense. The width is 13 feet, and the length is 13 + 2, or 15, feet. 4 Look Back The length of the room is 2 feet greater than the width. Also 13(15) = 195.

Lesson Quiz: Part I Complete the square to form a perfect square trinomial. 1. x2 +11x + 2. x2 – 18x + Solve by completing the square. 3. x2 – 2x – 1 = 0 4. 3x2 + 6x = 144 5. 4x2 + 44x = 23 81 6, –8

Lesson Quiz: Part II 6. Dymond is painting a rectangular banner for a football game. She has enough paint to cover 120 ft2. She wants the length of the banner to be 7 ft longer than the width. What dimensions should Dymond use for the banner? 8 feet by 15 feet