# Chapter 1 - Equations and Inequalities - 1.4 Quadratic Equations - 1.4 Exercises: 13

$\left\{2, 3\right\}$

#### 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 given trinomial has $b=-5$ and $c=6$. Note that $6=-2(-3)$ and $-5= -2+(-3)$. This means that $d=-2$ and $e=-3$ Thus, the factored form of the trinomial is: $[x+(-2)][x+(-3)] = (x-2)(x-3)$ The given equation maybe written as: $(x-2)(x-3)=0$ Use the Zero-Factor Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &x-2 = 0 &\text{ or } &x-3=0 \\&x=2 &\text{ or } &x=3 \end{array}$ Thus, the solution set is $\left\{2, 3\right\}$.

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