Question

The sum of all the real roots of the equation \[{{\left| x-2 \right|}^{2}}+\left| x-2 \right|-2=0\] is:

Answer 1

Hint: We will make the given equation a perfect square by adding, subtracting, multiplying and dividing the expression by any number such that it won’t change the given equation.
Complete step-by-step answer:
We have been given the equation \[{{\left| x-2 \right|}^{2}}+\left| x-2 \right|-2=0\].
In order to make the given equation a perfect square, we will write the equation in the form as shown below.
\[\begin{align}
  & {{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}} \\
 & \Rightarrow {{\left| x-2 \right|}^{2}}+\left| x-2 \right|-2=0 \\
 & ={{\left| x-2 \right|}^{2}}+2.\dfrac{1}{2}\left| x-2 \right|+{{\left( \dfrac{1}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}}-2=0 \\
 & ={{\left| x-2 \right|}^{2}}+2.\dfrac{1}{2}\left| x-2 \right|+{{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{1}{4}-2=0 \\
 & ={{\left( \left| x-2 \right|+\dfrac{1}{2} \right)}^{2}}-\dfrac{9}{4}=0 \\
 & ={{\left( \left| x-2 \right|+\dfrac{1}{2} \right)}^{2}}=\dfrac{9}{4} \\
\end{align}\]
Taking square root on both sides of the equation, we get,
\[\left( \left| x-2 \right|+\dfrac{1}{2} \right)=\pm \dfrac{3}{2}\]
Let us consider each root as a separate case and then try to find the value of x. For \[\left| x-2 \right|+\dfrac{1}{2}=\dfrac{3}{2}\], we get,
\[\begin{align}
  & \left| x-2 \right|+\dfrac{1}{2}=\dfrac{3}{2} \\
 & \left| x-2 \right|=\dfrac{3}{2}-\dfrac{1}{2} \\
 & \left| x-2 \right|=\dfrac{2}{2}=1 \\
 & \left| x-2 \right|=1 \\
 & x-2=\pm 1 \\
\end{align}\]
Hence, we get the values of \[x=3\] and \[x=1\].
For the case when \[\left| x-2 \right|+\dfrac{1}{2}=\dfrac{-3}{2}\], we get,
\[\begin{align}
  & \left| x-2 \right|+\dfrac{1}{2}=\dfrac{-3}{2} \\ .
 & \left| x-2 \right|=\dfrac{-3}{2}-\dfrac{1}{2} \\
 & \left| x-2 \right|=-2 \\
\end{align}\]
But we know that the value of the modulus function cannot be less than zero. So there is no solution at all for this case.
Thus the only real roots are 3 and 1. We have been asked to find the sum of real roots in the question.
Hence, sum of roots = 3 + 1 = 4.
Hence, the sum of real roots of the given equation is equal to 4.
Note: The important step here is to be able to express the given equation in terms of a perfect square. The conversion must be done carefully, avoiding any silly mistakes and also avoiding missing out some terms. Remember the point that a modulus function is always open with \[\pm \] sign, here we don’t get any root for negative value but don’t skip it anywhere.