Question

Find the number of solutions of ${{x}^{2}}+|x-1|=1$.

Answer 1

We here have been given the ${{x}^{2}}+|x-1|=1$ and we need to find the number of its solutions. For this, we will first open the modulus function in this equation which opens up as |y|=\left\\{ \begin{matrix} & y &y;\ge 0 \\\ & -y &y;<0 \\\ \end{matrix} \right.. Then, there will be two different equations formed in two different domains. We will solve both of those equations and we will only consider those solutions which lie in the respective domain. After doing that for both the equations, we will count the total number of solutions for equations and hence we will get our answer.

Complete step by step answer:
Here, we need to find the number of solutions to the equation ${{x}^{2}}+|x-1|=1$. For this, we will have to open up the modulus function first.
We know that the modulus function opens as follows:
|y|=\left\\{ \begin{matrix} & y &y;\ge 0 \\\ & -y &y;<0 \\\ \end{matrix} \right.
Thus, if we take the LHS of the equation to be f(x), we will get:
$f\left( x \right)={{x}^{2}}+|x-1|$
This will open as:
f\left( x \right)=\left\\{ \begin{matrix} & {{x}^{2}}+\left( x-1 \right) &x-1;\ge 0 \\\ & {{x}^{2}}-\left( x-1 \right) &x-1;<0 \\\ \end{matrix} \right.
Hence, we will get:
f\left( x \right)=\left\\{ \begin{matrix} & {{x}^{2}}+x-1 &x;\ge 1 \\\ & {{x}^{2}}-x+1 &x;<1 \\\ \end{matrix} \right.
Since the RHS of this equation is given to us as 1, we need to find the solutions to $f\left( x \right)=1$.
We will find this in both the different domains.
Case-1: $x\ge 1$
The equation is:
${{x}^{2}}+x-1=1$
Solving this by factorization method we get:
$\begin{aligned} & {{x}^{2}}+x-2=0 \\\ & \Rightarrow {{x}^{2}}+2x-x-2=0 \\\ & \Rightarrow x\left( x+2 \right)-\left( x+2 \right)=0 \\\ & \Rightarrow \left( x-1 \right)\left( x+2 \right)=0 \\\ \end{aligned}$
Thus, we get the value of x as:
$\begin{aligned} & x-1=0 \\\ & \therefore x=1 \\\ & x+2=0 \\\ & \therefore x=-2 \\\ \end{aligned}$
Hence, the values for x are 1 and -2. But here, the value of x is always either greater than or equal to 1. Hence, x cannot be equal to -2.
Hence, here x=1 …..(i)
Case-2: $x<1$
The equation is:
${{x}^{2}}-x+1=1$
Solving it by factorization method we get:
$\begin{aligned} & {{x}^{2}}-x+1=1 \\\ & \Rightarrow {{x}^{2}}-x=0 \\\ & \Rightarrow x\left( x-1 \right)=0 \\\ \end{aligned}$
Thu, we get the value of x as:
$\begin{aligned} & \therefore x=0 \\\ & x-1=0 \\\ & \therefore x=1 \\\ \end{aligned}$
Hence, the values for x are 1 and 0. But here, the value of x is always less than 1. Hence, x cannot be equal to 1.
Hence, here x=0 …..(ii)
From equations (i) and (ii), we can see that x=0,1.

Hence, there are 2 solutions to the given equation.

Note: Always solve the questions which involve the modulus function by opening the modulus function and then solving them in their different domains. Keeping in mind the domains is the most important part of these types of questions. This way, the scope of committing mistakes will be at its lowest.