Question

Find the number of real solution to the rational equation
$\frac{(2x+1)}{(x+2)}=x-1$

Answer 1

Solve the given equation as follows:

$\frac{(2x+1)}{(x+2)}=x-1$

$2x+1=(x-1)(x+2)$

$2x+1=x^2+x-2$

$x^2+x-2-2x-1=0$

$x^2-x-3=0$

Now we find the roots of above quadratic equation as shown below:

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

$=\frac{-(-1)\pm \sqrt{{(-1)}^2-4\left(1\right)(-3)}}{2\left(1\right)}$

$=\frac{1\pm \sqrt{1+12}}{2}$

$=\frac{1\pm \sqrt{13}}{2}$

Therefore, $x=\frac{1\pm \sqrt{13}}{2}$ .