# Chapter 1 - Section 1.4 - Quadratic Equations - 1.4 Exercises - Page 112: 13

$x=\{ 2, 3 \}$

#### Work Step by Step

The two numbers whose product is $ac=1(6)=6$ and whose sum is $b= -5$ are $\{ -2,-3 \}$. Using these two numbers to decompose the middle term of the given equation, $x^2-5x+6=0 ,$ results to \begin{array}{l}\require{cancel} x^2-2x-3x+6=0 \\\\ (x^2-2x)-(3x-6)=0 \\\\ x(x-2)-3(x-2)=0 \\\\ (x-2)(x-3)=0 .\end{array} Equating each factor to zero (or the Zero-Factor Property), the solutions to the given equation are \begin{array}{l}\require{cancel} x-2=0 \\\\ x=2 ,\\\\\text{OR}\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solutions are $x=\{ 2, 3 \} .$

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