# The number of the real roots of the equation

Question:

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

Solution:

$x \geq 5$

$(x+1)^{2}+(x-5)=\frac{27}{4}$

$\Rightarrow x^{2}+3 x-4=\frac{27}{4}$

$\Rightarrow x^{2}+3 x-\frac{43}{4}=0$

$\Rightarrow 4 x^{2}+12 x-43=0$

$\boldsymbol{X}=\frac{-12 \pm \sqrt{144+688}}{8}$

$x=\frac{-12 \pm \sqrt{832}}{8}=\frac{-12 \pm 28.8}{8}$

$=\frac{-3 \pm 7.2}{2}$

$=\frac{-3+7.2}{2}, \frac{-3-7.2}{2}$ (Therefore no solution)

For $x \leq 5$

$(x+1)^{2}-(x-5)=\frac{27}{4}$

$x^{2}+x+6-\frac{27}{4}=0$

$x=\frac{-4 \pm \sqrt{16+48}}{8}$

$x=\frac{-4 \pm 8}{8} \Rightarrow x=-\frac{12}{8}, \frac{4}{8}$

$\therefore 2$ Real Root's