## Intermediate Algebra (12th Edition)

$x=3$
$\bf{\text{Solution Outline:}}$ Express the given equation, $-x^2=9-6x ,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation. $\bf{\text{Solution Details:}}$ The factored form of the equation above is \begin{array}{l}\require{cancel} -x^2+6x-9=0 \\\\ -1(-x^2+6x-9)=0(-1) \\\\ x^2-6x+9=0 \\\\ (x-3)(x-3)=0 .\end{array} Equating each factor to zero (Zero-Factor Property), then \begin{array}{l}\require{cancel} x-3=0 \text{ OR } x-3=0 .\end{array} Using the properties of equality to solve each of the equation above results to \begin{array}{l}\require{cancel} x-3=0 \\\\ x=3 \\\\\text{ OR }\\\\ x-3=0 \\\\ x=3 .\end{array} Hence, the solution is $x=3 .$ $\bf{\text{Supplementary Solution/s:}}$ To factor the expression, $x^2-6x+9 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the expression above, the value of $c$ is $9$ and the value of $b$ is $-6 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,9 \}, \{ 3,3 \}, \{ -1,-9 \}, \{ -3,-3 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,-3 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x-3)(x-3) .\end{array}