Intermediate Algebra (6th Edition)

Published by Pearson

Chapter 8 - Section 8.2 - Solving Quadratic Equations by the Quadratic Formula - Exercise Set - Page 494: 78

Answer

The solution set is $\left\{-12, 1\right\}$. This means that both $-12$ and $1$ satisfy the equation.

Work Step by Step

RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The trinomial in the given equation has $b=11$ and $c=-12$. Note that $-12=12(-1)$ and $11 = 12+(-1)$. This means that $d=12$ and $e=-1$ Thus, the factored form of the trinomial is: $(p+12)[p+(-1)]= (p+12)(p-1)$ The given equation can be written as: $(p+12)(p-1)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &p+12 = 0 &\text{ or } &p-1=0 \\&p=-12 &\text{ or } &p=1 \end{array}$ Thus, the solution set is $\left\{-12, 1\right\}$. This means that both $-12$ and $1$ satisfy the equation.

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