# The number of solutions of x

Question:

The number of solutions of $x^{2}+|x-1|=1$ is

(a) 0

(b) 1

(c) 2

(d) 3

Solution:

(c) 2

$x^{2}+|x-1|=x^{2}+x-1, x \geq 1$

$=x^{2}-x+1 \quad, x<1$

(i) $x^{2}+x-1=1$

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

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

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow(x+2)(x-1)=0$

$\Rightarrow x+2=0$ or, $x-1=0$

$\Rightarrow x=-2$ or $x=1$

Since $-2$ does not satisfy the condition $x \geq 1$

(ii) $x^{2}-x+1=1$

$\Rightarrow x^{2}-x=0$

$\Rightarrow x^{2}-x=0$

$\Rightarrow x(x-1)=0$

$\Rightarrow x=0$ or, $(x-1)=0$

$\Rightarrow x=0, \quad x=1$

$x=1$ does not satisfy the condition $x<1$

So, there are two solutions.