§ 6.6 Solving Quadratic Equations by Factoring. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Zero Factor Theorem Quadratic Equations Can be.

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§ 6.6 Solving Quadratic Equations by Factoring

Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 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 a and b are real numbers and ab = 0, then a = 0 or b = 0. This theorem is very useful in solving quadratic equations.

Martin-Gay, Beginning and Intermediate Algebra, 4ed 33 Steps for Solving a Quadratic Equation by Factoring 1)Write the equation in standard form so that one side of the equation is 0. 2)Factor the quadratic expression completely. 3)Set each factor containing a variable equal to 0. 4)Solve the resulting equations. 5)Check each solution in the original equation. Solving Quadratic Equations

Martin-Gay, Beginning and Intermediate Algebra, 4ed 44 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 Solving Quadratic Equations Continued. Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 55 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. Solving Quadratic Equations Example continued:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 66 Solve 4x(8x + 9) = 5 First write the quadratic equation in standard form. 32x x = 5 32x x – 5 = 0 Now we factor the quadratic using techniques from the previous sections. 32x x – 5 = (8x – 1)(4x + 5) = 0 We set each factor equal to 0. 8x – 1 = 0 or 4x + 5 = 0 Solving Quadratic Equations Continued. 8x = 1 or 4x = – 5, which simplifies to x = or Example:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 77 Check both possible answers in the original equation. true So our solutions for x are or. Solving Quadratic Equations Example continued:

Martin-Gay, Beginning and Intermediate Algebra, 4ed 88 Recall that in Chapter 3, we found the x-intercept of linear equations by letting y = 0 and solving for x. The same method works for x-intercepts in quadratic equations. Note: When the quadratic equation is written in standard form, the graph is a parabola opening up (when a > 0) or down (when a < 0), where a is the coefficient of the x 2 term. The intercepts will be where the parabola crosses the x-axis. Finding x-intercepts

Martin-Gay, Beginning and Intermediate Algebra, 4ed 99 Find the x-intercepts of the graph of y = 4x x + 6. The equation is already written in standard form, so we let y = 0, then factor the quadratic in x. 0 = 4x x + 6 = (4x + 3)(x + 2) We set each factor equal to 0 and solve for x. 4x + 3 = 0 or x + 2 = 0 4x = – 3 or x = – 2 x = – ¾ or x = – 2 So the x-intercepts are the points ( – ¾, 0) and ( – 2, 0). Finding x-intercepts Example: