Question

In the following exercises, solve equations in Quadratic Form for $6x^{-2}\hspace{0.17em}+\hspace{0.17em}13x^{-1}\hspace{0.17em}+\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}0$.

Short answer

The solution of the equation is $x=-\frac{3}{5},\hspace{0.17em}x=-2$.

Step-by-step solution

Step 1. Given Information.

The given equation is $6x^{-2}\hspace{0.17em}+\hspace{0.17em}13x^{-1}\hspace{0.17em}+\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}0$.

Step 2. Simplify

Simplifying $6x^{-2}\hspace{0.17em}+\hspace{0.17em}13x^{-1}\hspace{0.17em}+\hspace{0.17em}5\hspace{0.17em}=\hspace{0.17em}0$

$$\begin{gathered} \frac{6}{x^2}+\frac{13}{x}+5=0 \\ 6+13x+5x^2=0 \end{gathered}$$

Step 3. Solve Quadratic equation.

Using quadratic formula:-

$5x^2+13x+6=0$

$$\begin{gathered} x_{1,\hspace{0.17em}2}=\frac{-13\pm \sqrt{{13}^2-4·\hspace{0.17em}5·\hspace{0.17em}6}}{2·\hspace{0.17em}5} \\ {x}_{1,\hspace{0.17em}2}=\frac{-13\pm \hspace{0.17em}7}{2·\hspace{0.17em}5} \\ {x}_1=\frac{-13+7}{2·\hspace{0.17em}5},\hspace{0.17em}{x}_2=\frac{-13-7}{2·\hspace{0.17em}5} \\ x=-\frac{3}{5},\hspace{0.17em}x=-2 \end{gathered}$$

Therefore, The solution of the equation are $x=-\frac{3}{5},\hspace{0.17em}x=-2$.