Chapter 5: Exercises with Solutions

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476 Chapter 5 Quadratic FunctionsIn Exercises 61-66, Use the graph off(x) = ax 2 + bx + c shown to find all solutionsof the equation f(x) = 0. (Note:Every solution is an integer.)64.y561.y5x5x565.y562.y5x5x566.y563.y5x5x5Version: Fall 2007
Section 5.3Zeros of the Quadratic5.3 Solutions1. Look for p and q such that (x+p)(x+q) = x 2 +9x+14. It follows that p+q = 9 andpq = 14, so p = 2 and q = 7. Now verify the factorization by multiplying (x + 2)(x + 7)to obtain x 2 + 9x + 14.3. Look for p and q such that (x+p)(x+q) = x 2 +10x+9. It follows that p+q = 10 andpq = 9, so p = 9 and q = 1. Now verify the factorization by multiplying (x + 9)(x + 1)to obtain x 2 + 10x + 9.5. Look for p and q such that (x + p)(x + q) = x 2 − 4x − 5. It follows that p + q = −4and pq = −5, so p = −5 and q = 1. Now verify the factorization by multiplying(x − 5)(x + 1) to obtain x 2 − 4x − 5.7. Look for p and q such that (x + p)(x + q) = x 2 − 7x + 12. It follows that p + q = −7and pq = 12, so p = −4 and q = −3. Now verify the factorization by multiplying(x − 4)(x − 3) to obtain x 2 − 7x + 12.9. To find the zeroes, set f(x) = 0 and factor.By the zero product property, either0 = x 2 − 2x − 15Solve these linear equations independently.0 = (x − 5)(x + 3)x − 5 = 0 or x + 3 = 0.x = 5 or x = −311. To find the zeroes, set f(x) = 0 and factor.0 = x 2 + 10x − 390 = (x + 13)(x − 3)By the zero product property, eitherx + 13 = 0 or x − 3 = 0.Solve these linear equations independently.x = −13 or x = 3Version: Fall 2007
Page 1 and 2: Section 5.1 The Parabola 4335.1 Exe
Page 3 and 4: Section 5.1 The Parabola 435In Exer
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Chapter 5: Exercises with Solutions

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