Question

Find the real solutions of the quadratic equation $\frac{10}{x}-\frac{12}{x-3}+4=0$.

Short answer

The given quadratic has two real solutions $x=5,-\frac{3}{2}$.

Step-by-step solution

Step 1. By method of quadratic formula.

The roots of the quadratic equation of the form $a{x}^2+bx+c=0$where $a\ne 0$can be evaluated by using the quadratic formula shown below.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 2. Find roots of quadratic equation by quadratic formula.

The given equation is$\frac{10}{x}-\frac{12}{x-3}+4=0$ which can be written as,

role="math" localid="1643834311479" $$\begin{gathered} \frac{10}{x}-\frac{12}{x-3}+4=0....\left(\text{Given}\right) \\ \text{or,\hspace{0.17em}}10\left(x-3\right)-12x+4x\left(x-3\right)=0....\left(\text{Simplify}\right) \\ 10x-30-12x+4x^2-12x=0 \\ 4x^2-14x-30=0....\left(1\right) \end{gathered}$$

Evaluate roots of (1) by quadratic formula as shown below.

$$\begin{gathered} x=\frac{-\left(-14\right)\pm \sqrt{{\left(-14\right)}^2-4\left(4\right)\left(-30\right)}}{2\left(4\right)} \\ =\frac{14\pm \sqrt{676}}{8} \\ =\frac{14\pm 26}{8} \\ =5,-\frac{3}{2} \end{gathered}$$

Step 3. Description of steps.

The given quadratic equation has two distinct real solutions $x=5,-\frac{3}{2}$.