Question

How do you solve $ {(x - 1)^2} = 9 $ ?

Answer 1

Hint: In such questions, try to simplify the equation by eliminating higher degree terms. Here, LHS is a quadratic equation and a perfect square. We can also see that the RHS can also be written as a square of a number so square rooting both the sides will simplify the LHS and it will become a linear equation which would be easier to solve.

Complete step-by-step answer:
(i)
We are given,
  $ {(x - 1)^2} = 9 $
As we know that $ 9 $ is a perfect square of $ 3 $ , square rooting both the sides will give us a simpler equation:
  $ \sqrt {{{(x - 1)}^2}} = \sqrt 9 $
In LHS, the square root will be cut with the square of $ (x - 1) $ and we will get a linear polynomial $ (x - 1) $ and in RHS $ \sqrt 9 $ will give us $ \pm 3 $ as squaring both $ + 3 $ and $ - 3 $ gives us $ 9 $ as a result.
Therefore, we get:
  $ (x - 1) = \pm 3 $
(ii)
Separating the equation in two different equations, we get:
  $ (x - 1) = 3 $ as first equation
And,
  $ (x - 1) = - 3 $ as the second equation
(iii)
Solving the first equation,
  $ x - 1 = 3 $
Keeping the variable $ x $ in LHS and shifting everything else to RHS will give us,
  $ x = 3 + 1 $ (because when $ - 1 $ got shifted from LHS to RHS, it changes it sign from negative to positive)
We get,
  $ x = 4 $
(iv)
Solving the second equation,
  $ x - 1 = - 3 $
Again, keeping the variable $ x $ in LHS and shifting everything else to RHS will give us,
  $ x = - 3 + 1 $
Since we know that when a negative sign and a positive sign comes together, we subtract and put the sign of the larger number. Here since, $ 3 > 1 $ we will put a negative sign.
Thus, we get:
  $ x = - 2 $
Hence, the solution of $ {(x - 1)^2} = 9 $ is $ x = 4 $ and $ x = - 2 $
So, the correct answer is “ $ x = 4 $ and $ x = - 2 $ ”.

Note: Always remember that there are two square-roots of a number. One, positive and the other, its negative value. For example, square-root of $ 9 $ i.e., $ \sqrt 9 $ is $ + 3 $ and $ - 3 $ as square of both of them yields us with $ 9 $ . Most of the time, students forget about $ - 3 $ and write $ \sqrt 9 $ directly equal to $ + 3 $ only which leads them to incorrect solutions. Also, if you don’t get a perfect square in RHS, you can also apply the identity $ ({a^2} + {b^2}) = {a^2} + 2ab + {b^2} $ to open the parentheses in LHS and solve the quadratic equation formed with the formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .