Question

The number of solution of the equation ${\left|x\right|}^2+|x-3|+a=0$; where $a=\{\begin{array}{c}-3,ifx\ge 3\\ -\frac{11}{4},ifx<3\end{array}$ is -

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Case I :

$x\ge 3$

$x^2+x-3-3=0$

$x^2+x-6=0$

$x^2+3x-2x-6=0$

$x(x+3)-2(x+3)=0$

$x=2,x=-3$

Case II:

$0\le x<3$

$x^2+3-x-\frac{11}{4}=0$

$4x^2+12-4x+1=0$

$(2x-1{)}^2=0$

$x=1/2$

Case III:

$x<0$

$x^2+3-x-11/4=0$

$x=1/2$

Only $1$ Solution