Question

The sum of all the real numbers satisfying the equation $x^2+|x-3|=4$ is

• A. 0
• B. 1
• C. 2
• D. $-1$

Answer 1

Answer: (B)

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
$\frac{1+\sqrt{5}}{2}+\frac{1-\sqrt{5}}{2}=1$