## Intermediate Algebra for College Students (7th Edition)

Solution set: $\{4,10\}$
$g(a)=7$ $\displaystyle \frac{a}{2}+\frac{20}{a}=7$ Exclude solutions which yield 0 in denominators: $a\neq 0\qquad (*)$ Multiply the equation with the LCD=$2a$ $a^{2}+40=14a$ $a^{2}-14a+40=0$ ... $(x^{2}+bx+c$ is factored by searching for factors of c whose sum is b). ... -10 and -4 are factors of 40 whose sum is -14 $(a-10)(a-4)=0$ By the zero product principle, $a\in\{4,10\},$ ... both values satisfy (*). Solution set: $\{4,10\}$