Question

Find all real solutions to the equation with absolute value of $ \left| {x - 1} \right| = \left| {{x^2} - 2x + 1} \right| $
A.No solutions
B. $ 0 $
C. $ 3 $
D. $ 0,1,2 $
D. $ 2 + \sqrt 2 ,{\text{ 2 - }}\sqrt 2 $

Answer 1

Hint: Here first of all we will find two separate equations considering one side equal to positive of the other side of the given equation and second time negative of the given equation. Then using the basic mathematical concept find the value of “x”.

Complete step-by-step answer:
Take the given expression –
Find all real solutions to the equation with absolute value
 $ \left| {x - 1} \right| = \left| {{x^2} - 2x + 1} \right| $
It can be re-written as – Moving the left hand side of the equation to right hand side of the equation and vice-versa.
 $ \left| {{x^2} - 2x + 1} \right| = \left| {x - 1} \right| $
To find the real solutions remove mode and take plus or minus sign on the right hand side of the equation.
 $ \left( {{x^2} - 2x + 1} \right) = \pm \left( {x - 1} \right) $
Take one the left hand side of the equation to the negative of the right hand side of the equation and the second time positive to the right hand side of the equation.
 $ \left( {{x^2} - 2x + 1} \right) = + \left( {x - 1} \right) $
Open the brackets; remember when there is a positive sign outside the bracket then values inside the bracket does not change.
 $ \Rightarrow {x^2} - 2x + 1 = x - 1 $
Take all the terms on one side of the equation. Also, when any term is moved from one side to another then the sign also changes. Positive term changes to negative and vice-versa.
 $ \Rightarrow {x^2} - 2x + 1 - x + 1 = 0 $
Make a pair of like terms.
 $ \Rightarrow {x^2}\underline { - 2x - x} + \underline {1 + 1} = 0 $
Simplify the above equation –
 $ \Rightarrow {x^2} - 3x + 2 = 0 $
By factorization method, the equation can be re-written as –
 $ \Rightarrow \underline {{x^2} - 2x} - \underline {x + 2} = 0 $
Take out the common multiple in the two pairs-
 $ \Rightarrow x(x - 2) - 1(x - 2) = 0 $
Take common multiple out –
 $ \Rightarrow (x - 2)(x - 1) = 0 $
We get –
 $
   \Rightarrow x - 1 = 0\,{\text{ or x - 2 = 0}} \\
   \Rightarrow {\text{x = 1 or x = 2}}\;{\text{ }}.....{\text{ (a)}} \\
  $
Similarly for the second case-
 $ \left( {{x^2} - 2x + 1} \right) = - \left( {x - 1} \right) $
Open the brackets; remember when there is a negative sign outside the bracket then values inside the bracket do change. Positive term changes to negative and vice-versa.
 $ \Rightarrow {x^2} - 2x + 1 = - x + 1 $
Take all the terms on one side of the equation. Also, when any term is moved from one side to another then the sign also changes. Positive term changes to negative and vice-versa.
 $ \Rightarrow {x^2} - 2x + 1 + x - 1 = 0 $
Make a pair of like terms.
 $ \Rightarrow {x^2}\underline { - 2x + x} + \underline {1 - 1} = 0 $
Simplify the above equation –
 $ \Rightarrow {x^2} - x = 0 $
Take out the common multiple in the above equation -
 $ \Rightarrow x(x - 1) = 0 $
 We get –
 $
   \Rightarrow x - 1 = 0\,{\text{ or }}\;{\text{x = 0}} \\
   \Rightarrow {\text{x = 1 or x = 0}}\;{\text{ }}.....{\text{ (b)}} \;
  $
Therefore, the real solutions of the given equation are $ 0,1{\text{ and 2}} $
So, the correct answer is “Option 4”.

Note: Always be sure when you move one term from one side of the equation to another. The sign of the term always changes when it is moved from one side to another in the equation. Positive term changes to negative and negative term changes to positive.