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The sum of the roots of the equation, ${x^2}\, + \,\left| {2x - 3} \right|\, - \,4\, = \,0,$ is
$2$
$-2$
$\sqrt 2$
$-\sqrt 2$
Solution
${x^2} + |2x – 3| – 4 = 0$
$|2 x-3|=\left\{\begin{array}{ccc}{(2 x-3)} & {\text { if }} & {x>\frac{3}{2}} \\ {-(2 x-3)} & {\text { if }} & {x<\frac{3}{2}}\end{array}\right.$
for $x>\frac{3}{2},$
$x^{2}+2 x-3-4=0$
$x^{2}+2 x-7=0$
$x=\frac{-2 \pm \sqrt{4+28}}{2}$
$ = \frac{{ – 2 \pm 4\sqrt 2 }}{2} = – 1 \pm 2\sqrt 2 $
Here $x=2 \sqrt{2}-1$
$\left\{2 \sqrt{2}-1<\frac{3}{2}\right\}$
for $x<\frac{3}{2}$
$x^{2}-2 x+3-4=0$
$\Rightarrow x^{2}-2 x-1=0$
$ \Rightarrow x = \frac{{2 \pm \sqrt {4 + 4} }}{2}$
$= \frac{{2 \pm 2\sqrt 2 }}{2}$
$= 1 \pm \sqrt 2 $
Here $x=1-\sqrt{2} \quad\left\{(1-\sqrt{2})<\frac{3}{2}\right\}$
Sum of roots : $(2 \sqrt{2}-1)+(1-\sqrt{2})=\sqrt{2}$
