Question

In the following exercises, solve.

$15x^{-2}-26x^{-1}+8=0$

Short answer

The solutions of the given equation are $x=\frac{5}{2}$and localid="1644991122921" $x=\frac{3}{4}$.

Step-by-step solution

Step 1. Write the equation in standard form.

The given equation is:

$15x^{-2}-26x^{-1}+8=0$

It can be written as:

$15(x^{-1}{)}^2-26x^{-1}+8=0$

Substitute $x^{-1}=u$.

$15u^2-26u+8=0$

Step 2. Solve the equation.

Split the middle term.

$\begin{array}{rcl}15u^2-20u-6u+8& =& 0\\ 5u(3u-4)-2(3u-4)& =& 0\\ (5u-2)(3u-4)& =& 0\\ u& =& \frac{2}{5}\text{and}u=\frac{4}{3}\end{array}$

Substitute $u=x^{-1}$.

localid="1644991105556" $\begin{array}{rcl}{x}^{-1}& =& \frac{2}{5}\text{and}{x}^{-1}=\frac{4}{3}\\ x& =& \frac{5}{2}\text{and}x=\frac{3}{4}\end{array}$

Step 3. Conclusion.

The solutions of the given equation are $x=\frac{5}{2}$and localid="1644991088222" $x=\frac{3}{4}$.