## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The solutions are $x=-14$ or $x=0$.
$g(x)=x^{2}+14x+49$ $g(x)=49\qquad$...substitute $49$ for $g(x)$. $49=x^{2}+14x+49$ Apply Perfect square formula: $(x+a)^{2}=x^{2}+2ax+a^{2}$, $49=(x+7)^{2}$ According to the general principle of square roots: For any real number $k$ and any algebraic expression $x$ : $\text{If }x^{2}=k,\text{ then }x=\sqrt{k}\text{ or }x=-\sqrt{k}$. $x+7=\pm\sqrt{49}\qquad$...add $-7$ to each side. $x+7-7=\pm\sqrt{49}-7\qquad$...simplify. $x=-7\pm 7$ $x=0$ or $x=-14$