Question
Solve the following equation by using formula :$x-\frac{1}{x}=3, x \neq 0$
✓
Answer
$
\begin{aligned}
& x-\frac{1}{x}=3 \\
& x^2-1=3 x \\
& \Rightarrow x^2-3 x-1=0
\end{aligned}
$
Here $a=1, b=-3, c=-1$
$
\begin{aligned}
\therefore & b^2-4 a c \\
= & (-3)^2-4 \times 1 \times(-1) \\
= & 9+4 \\
= & 13
\end{aligned}
$
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-(-3) \pm \sqrt{13}}{2 \times 1} \\
& =\frac{3 \pm \sqrt{13}}{2}
\end{aligned}
$
$\therefore x=\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.
\begin{aligned}
& x-\frac{1}{x}=3 \\
& x^2-1=3 x \\
& \Rightarrow x^2-3 x-1=0
\end{aligned}
$
Here $a=1, b=-3, c=-1$
$
\begin{aligned}
\therefore & b^2-4 a c \\
= & (-3)^2-4 \times 1 \times(-1) \\
= & 9+4 \\
= & 13
\end{aligned}
$
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{-(-3) \pm \sqrt{13}}{2 \times 1} \\
& =\frac{3 \pm \sqrt{13}}{2}
\end{aligned}
$
$\therefore x=\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.
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