Chapter A.6, Problem 76AYU

To determine

To solve: $x+\frac{12}{x}=7$

Answer to Problem 76AYU

$x=3, 4$

Explanation of Solution

$x+\frac{12}{x}=7$

$x^2+12=7x$   (Multiply $x$ on both sides)

$x^2-7x+12=0$   (Subtract $7x$ on both sides)

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=1, b= -7$, and $c=12$, the discriminant is $b^2-4ac={\left(-7\right)}^2-4\left(1\right)\left(12\right)=1$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=1, b= -7, c=12$ and $b^2-4ac=1$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-7\right)\pm \sqrt{1}}{2\left(1\right)}=\frac{7\pm 1}{2}$

$x=\frac{7+1}{2}=4$ and $x=\frac{7-1}{2}=3$

Therefore $x=3, 4$.