Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 4 - Quadratic Functions - 4.5 Solving Equations by Factoring - 4.5 Exercises - Page 362: 89


The solution set is $\left\{2, 5\right\}$.

Work Step by Step

Recall: If the trinomial $x^2+bx+c$ is factorable, then its factored form is $(x+d)(x+e)$ where $c=de$ and $b=d+e$. To check if the trinomial is factorable, look for factors of $10$ whose sum is equal to $-7$. Note that $10=-5(-2)$ and $-7=-5+(-2)$. Thus, $d=-5$ and $e=-2$, and the trinomial is factorable. Hence, $$p^2-7p+10=(p-5)(p-2)$$ The given equation is equivalent to $(p-5)(p-2)=0$. Use the Zero-Product Property by equating each factor to zero, then solving each equation to obtain: \begin{align*} p-5&=0 &\text{or}& &p-2=0\\ p&=5 &\text{or}& &p=2\\ \end{align*} Checking: \begin{align*} 5^2-7(5)+10&=0\\ 25-35+10&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} \begin{align*} 2^2-7(2)+10&=0\\ 4-14+10&=0\\ 0&\stackrel{\checkmark}=0 \end{align*} Therefore, the solution set is $\left\{2, 5\right\}$.
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