Question

Find the roots of the following quadratic equations by using the quadratic formula
$\frac{x}{x-1}+\frac{x-1}{x}=4,x\ne 0,1$

• A. $\frac{2\pm \sqrt{3}}{4}$
• B. $\frac{-1\pm \sqrt{3}}{2}$
• C. $\frac{1\pm \sqrt{3}}{1}$
• D. $\frac{-2\pm \sqrt{3}}{4}$

Answer 1

The correct option is C $\frac{1\pm \sqrt{3}}{1}$

Given, $\frac{x}{x-1}+\frac{x-1}{x}=4$
$\Rightarrow \frac{x^2+(x-1{)}^2}{x(x-1)}=4$
$\Rightarrow x^2+x^2-2x+1=4x^2-4x$
$\Rightarrow 2x^2-4x^2-2x+4x+1=0$
$\Rightarrow -2x^2+2x+1=0$
$\Rightarrow 2x^2-2x-1=0$
Comparing $2x^2-2x-1=0$ with $a{x}^2+bx+c=0$, we get $a=2,b=-2.c=-1$
Now the roots are $=$ $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=$ $\frac{-(-2)\pm \sqrt{(-2{)}^2-4\times 2\times (-1)}}{2\times 1}$
$=$ $\frac{2\pm \sqrt{4+8}}{2}$
$=$ $\frac{2\pm \sqrt{12}}{2}$
$=$ $\frac{2\pm 2\sqrt{3}}{2}$
$=$ $1\pm \sqrt{3}$