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§ 3.6 Solving Quadratic Equations by Factoring. Martin-Gay, Developmental Mathematics 2 Zero Factor Theorem Quadratic Equations Can be written in the.

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Presentation on theme: "§ 3.6 Solving Quadratic Equations by Factoring. Martin-Gay, Developmental Mathematics 2 Zero Factor Theorem Quadratic Equations Can be written in the."— Presentation transcript:

1 § 3.6 Solving Quadratic Equations by Factoring

2 Martin-Gay, Developmental Mathematics 2 Zero Factor Theorem Quadratic Equations Can be written in the form ax 2 + bx + c = 0. a, b and c are real numbers and a  0. This is referred to as standard form. Zero Factor Theorem If m and n are real numbers and mn = 0, then m = 0 or n = 0. This theorem is very useful in solving quadratic equations.

3 Martin-Gay, Developmental Mathematics 3 Solving Quadratic Equations Steps for Solving a Quadratic Equation by Factoring 1)Write the equation in standard form. 2)Factor the quadratic completely. 3)Set each factor containing a variable equal to 0. 4)Solve the resulting equations. 5)Check each solution in the original equation.

4 Martin-Gay, Developmental Mathematics 4 Solving Quadratic Equations In general: ax 2 + bx + c = 0.

5 Martin-Gay, Developmental Mathematics 5 Solving Quadratic Equations Solve x 2 – 5x = 24. First write the quadratic equation in standard form. x 2 – 5x – 24 = 0 Now we factor the quadratic using techniques from the previous sections. x 2 – 5x – 24 = (x – 8)(x + 3) = 0 We set each factor equal to 0. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = – 3 Example 1 Continued.

6 Martin-Gay, Developmental Mathematics 6 Solving Quadratic Equations Check both possible answers in the original equation. 8 2 – 5(8) = 64 – 40 = 24 true (–3) 2 – 5(–3) = 9 – (–15) = 24 true So our solutions for x are 8 or –3. Example Continued

7 Martin-Gay, Developmental Mathematics 7 Solving Quadratic Equations Solve 4x(8x + 9) = 5 First write the quadratic equation in standard form. 32x 2 + 36x = 5 32x 2 + 36x – 5 = 0 Now we factor the quadratic using techniques from the previous sections. 32x 2 + 36x – 5 = (8x – 1)(4x + 5) = 0 We set each factor equal to 0. 8x – 1 = 0 or 4x + 5 = 0 Example 2 Continued. 8x = 1 or 4x = – 5, which simplifies to x = or

8 Martin-Gay, Developmental Mathematics 8 Solving Quadratic Equations Put it all together: 4x(8x + 9) = 5 32x 2 + 36x = 5 32x 2 + 36x – 5 = 0 (8x – 1)(4x + 5) = 0 8x – 1 = 0 or 4x + 5 = 0

9 Martin-Gay, Developmental Mathematics 9 Solving Quadratic Equations Check both possible answers in the original equation. true So our solutions for x are or. Example Continued


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