Introduction We can solve quadratic equations by transforming the left side of the equation into a perfect square trinomial and using square roots to solve. Previously, you may have explored perfect square trinomials in the form (ax) 2 ± 2abx + b 2. In this lesson, we will use a modified æ version of this form, x 2 + bx + b 2 ö è ç 2ø. 1
Key Concepts When the binomial (x + a) is squared, the resulting perfect square trinomial is x 2 + 2ax + a 2. When the binomial (ax + b) is squared, the resulting perfect square trinomial is a 2 x 2 + 2abx + b 2. 2
Key Concepts, continued Completing the Square to Solve Quadratics 1. Make sure the equation is in standard form, ax 2 + bx + c. 2. Subtract c from both sides. 3. Divide each term by a to get a leading coefficient of 1. 4. Add the square of half of the coefficient of the x-term to both sides to complete the square. 5. Express the perfect square trinomial as the square of a binomial. 6. Solve by using square roots. 3
Common Errors/Misconceptions neglecting to add the c term of the perfect square trinomial to both sides not isolating x after squaring both sides forgetting that, when taking the square root, both the positive and negative roots must be considered (±) 4
Guided Practice Example 2 Solve x 2 + 6x + 4 = 0 by completing the square. 5
Guided Practice: Example 2, continued 1. Determine if x 2 + 6x + 4 is a perfect square trinomial. Take half of the value of b and then square the result. If this is equal to the value of c, then the expression is a perfect square trinomial. æ ç è bö 2 ø 2 æ = 6 ö ç è 2 ø 2 = 9 x 2 + 6x + 4 is not a perfect square trinomial because the square of half of 6 is not 4. 6
Guided Practice: Example 2, continued 2. Complete the square. x 2 + 6x + 4 = 0 x 2 + 6x = 4 x 2 + 6x + 3 2 = 4 + 3 2 x 2 + 6x + 9 = 5 Original equation Subtract 4 from both sides. Add the square of half of the coefficient of the x-term to both sides to complete the square. Simplify. 7
Guided Practice: Example 2, continued 3. Express the perfect square trinomial as the square of a binomial. Half of b is 3, so the left side of the equation can be written as (x + 3) 2. (x + 3) 2 = 5 8
Guided Practice: Example 2, continued 4. Isolate x. (x + 3) 2 = 5 Equation x + 3 = ± 5 x = -3 ± 5 Take the square root of both sides. Subtract 3 from both sides. 9
Guided Practice: Example 2, continued 5. Determine the solution(s). The equation x 2 + 6x + 4 = 0 has two solutions, x = -3 ± 5. 10
Guided Practice: Example 2, continued 11
Guided Practice Example 3 Solve 5x 2 50x 120 = 0 by completing the square. 12
Guided Practice: Example 3, continued 1. Determine if 5x 2 50x 120 = 0 is a perfect square trinomial. The leading coefficient is not 1. First divide both sides of the equation by 5 so that a = 1. 5x 2 50x 120 = 0 Original equation x 2 10x 24 = 0 Divide both sides by 5. 13
Guided Practice: Example 3, continued Now that the leading coefficient is 1, take half of the value of b and then square the result. If the expression is equal to the value of c, then it is a perfect square trinomial. æ ç è bö 2 ø 2 æ = -10 ö ç è 2 ø 2 = 25 5x 2 50x 120 = 0 is not a perfect square trinomial because the square of half of 10 is not 24. 14
Guided Practice: Example 3, continued 2. Complete the square. x 2 10x 24 = 0 x 2 10x = 24 x 2 10x + ( 5) 2 = 24 + ( 5) 2 x 2 10x + 25 = 49 Equation Add 24 to both sides. Add the square of half of the coefficient of the x-term to both sides to complete the square. Simplify. 15
Guided Practice: Example 3, continued 3. Express the perfect square trinomial as the square of a binomial. Half of b is 5, so the left side of the equation can be written as (x 5) 2. (x 5) 2 = 49 16
Guided Practice: Example 3, continued 4. Isolate x. (x 5) 2 = 49 Equation x - 5 = ± 49 = ±7 x = 5 ± 7 x = 5 + 7 = 12 or x = 5 7 = 2 Take the square root of both sides. Add 5 to both sides. Split the answer into two separate equations and solve for x. 17
Guided Practice: Example 3, continued 5. Determine the solution(s). The equation 5x 2 50x 120 = 0 has two solutions, x = 2 or x = 12. 18
Guided Practice: Example 3, continued 19