Question

In each of the following, determine whether the given values are solutions of the given equation or not :
$x+\frac{1}{x}=\frac{1}{6},x=\frac{5}{6},x=\frac{4}{3}$

• A. $x$ = $\frac{5}{6}$ & $x$ = $\frac{4}{3}$ are not solutions.
• B. $x$ = $\frac{5}{6}$ & $x$ = $\frac{4}{3}$ are solutions.
• C. $x$ = $\frac{5}{6}$ is a solution but $x$ = $\frac{4}{3}$ not.
• D. $x$ = $\frac{4}{3}$ is a solution but $x$ = $\frac{5}{6}$ not.

Answer 1

The correct option is A $x$ = $\frac{5}{6}$ & $x$ = $\frac{4}{3}$ are not solutions.

$x+\frac{1}{x}=\frac{1}{6}$

$\Rightarrow$ $6x^2+6=x$

$\Rightarrow$ $6x^2-x+6=0$

$\Rightarrow$ Here, $a=6,b=-1,c=6$

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

$=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(6\left)\right(6)}}{2\left(6\right)}$

$=\frac{1\pm \sqrt{-143}}{12}$

$=\frac{1\pm \sqrt{143}i}{12}$

$\therefore$ $x=\frac{1+\sqrt{143}i}{12}$ and $x=\frac{1-\sqrt{143}i}{12}$