Chapter A.6, Problem 75AYU

To determine

To solve: $6x-5=\frac{6}{x}$

Answer to Problem 75AYU

$x=-\frac{2}{3},\frac{3}{2}$

Explanation of Solution

$6x-5=\frac{6}{x}$

$6x^2-5x=6$   (Multiply $x$ on both sides)

$6x^2-5x-6=0$   (Subtract 6 on both sides)

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

With $a=6, b= -5$, and $c=-6$, the discriminant is $b^2-4ac={\left(-5\right)}^2-4\left(6\right)\left(-6\right)=25+144=169$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=6, b= -5, c=-6$ and $b^2-4ac=169$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-5\right)\pm \sqrt{69}}{2\left(6\right)}=\frac{5\pm 13}{12}$

$x=\frac{5+13}{12}=\frac{18}{12}=\frac{3}{2}$ and $x=\frac{5-13}{12}=-\frac{8}{12}=-\frac{2}{3}$

Therefore $x=-\frac{2}{3},\frac{3}{2}$.