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Q.
The sum of all the real numbers satisfying the equation $x^{2}+|x-3|=4$ is
TS EAMCET 2019
Solution:
We have,
$x^{2}+|x-3|=4$ Case I. $x>3$
$\therefore x^{2}+x-3=4$
$x^{2}+x-7=0$
$x=\frac{-1 \pm \sqrt{29}}{2}$
$\therefore $ No solution $[\because x>3]$ Case II. $x<3$
$\therefore x^{2}-x+3=4$
$ x^{2}-x-1 =0 $
$ x =\frac{1 \pm \sqrt{5}}{2} $
$\therefore $ Sum of all real number satisfy ray the equal
$\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}=1$