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Algebra 1 Section 12.3.

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Presentation on theme: "Algebra 1 Section 12.3."— Presentation transcript:

1 Algebra 1 Section 12.3

2 Solving Quadratic Equations
Some quadratic equations cannot be solved by factoring. A method called “completing the square” can often be used when factoring cannot.

3 Completing the Square To complete the square of an expression in the form x2 + bx, add the square of half the coefficient of x: x2 + bx + (b/2)2.

4 Example 1a x2 – 16x Since b = -16, (b/2)2 = (-8)2 = 64 x2 – 16x + 64

5 Example 1b x2 + x Since b = 3/5, (b/2)2 = (3/10)2 = 9/100 x2 + x +

6 Completing the Square This method can be used to solve any quadratic equation of the form x2 + bx + c. Remember to add the same number to both sides of the equation.

7 Example 2 Solve x2 + 8x = 9. Since b = 8, (b/2)2 = (4)2 = 16
Add 16 to both sides. x2 + 8x + 16 = x2 + 8x + 16 = 25 (x + 4)2 = 25

8 Be sure to check your solutions in the original equation.
Example 2 (x + 4)2 = 25 x + 4 = ±5 x = -4 ± 5 x = 1, -9 Be sure to check your solutions in the original equation.

9 Solving Quadratic Equations by Completing the Square
Write the equation in the form x2 + bx = c. Complete the square by adding (b/2)2 to both sides of the equation. Factor the trinomial.

10 Solving Quadratic Equations by Completing the Square
Apply the Square Root Property: If x2 = n, then x = ± n . Solve the resulting equations.

11 Example 3 Solve a2 – 7a – 3 = 0. a2 – 7a = 3 (-7/2)2 = 49/4
Add 49/4 to both sides. a2 – 7a + 49/4 = /4 (a – 7/2)2 = 61/4

12 In decimal form, the solution set is ≈ {-0.41, 7.41}.
Example 3 (a – 7/2)2 = 61/4 a – = ± 61 2 7 a = 7 ± 61 2 In decimal form, the solution set is ≈ {-0.41, 7.41}.

13 Example 4 Let x = the original length and width of the square garden
x + 4 = the new width x + 7 = the new length Using the area formula: (x + 7)(x + 4) = 550

14 Example 4 (x + 7)(x + 4) = 550 x2 + 11x + 28 = 550 x2 + 11x = 522
121 4 (x )2 = 11 2 2209 4

15 Example 4 (x + )2 = x + = ± x + = ± x = 11 2 2209 4 11 2 2209 4 11 2
47 x = -11 ± 47 2

16 Example 4 x = -11 ± 47 2 x = 18 or -29 Since a negative answer is not reasonable, the original garden was 18 ft on each side.

17 Homework: pp


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