Question

The number of the real roots of the equation
$(x+1{)}^2+|x-5|=\frac{27}{4}$ is_______.

Answer 1

Answer: 2

Case-I
$x\le 5$
$(x+1{)}^2-(x-5)=\frac{27}{4}$
$(x+1{)}^2-(x+1)-\frac{3}{4}=0$
$x+1=\frac{3}{2},-\frac{1}{2}$
$x=\frac{1}{2},-\frac{3}{2}$
Case-II
$x>5$
$(x+1)+(x-5)=\frac{27}{4}$
$(x+1{)}^2+(x+1)-\frac{51}{4}=0$
$x=\frac{-1\pm \sqrt{52}}{2}($ rejected as $x>5)$
So, the equation have two real root.