Question

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

Answer 1

Hint: Since this is in the form of a polynomial equation, we will expand the equation and solve it by factorization to get the value of $x$. So, we will start by using ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ and expanding the LHS. Then, we will simplify and end up with a quadratic equation. We will use the factorisation method to get the values of x.

Complete answer:
We have the given term as:
${{(x-1)}^{2}}=9$
Since the left-hand side of the equation is in the form of ${{(a-b)}^{2}}$ we will expand it, we know that the value of ${{(a-b)}^{2}}$ can be expanded as:
${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
On expanding the equation, we get:
${{x}^{2}}-2x+1=9$.
Now on rearranging the equation we get:
${{x}^{2}}-2x+1-9=0$
On simplifying we get-
${{x}^{2}}-2x-8=0$
Since this equation is in the form of a quadratic equation, we will factorize the terms.
On splitting the middle term, we get:
${{x}^{2}}-4x+2x-8=0$
Now on taking common from both the sides, we get-
$x(x-4)+2(x-4)=0$
Since the term $(x-4)$ is common in both the terms, we take it out as common as-
$(x-4)(x+2)=0$
Therefore, either $x-4=0$ or $x+2=0$,
Therefore, either $x=4$or $x=-2$.
Therefore, the quadratic equation has $2$ roots which are $x=4$ and $x=-2$.

Note:
To check whether the answer is correct a student must put the value of the roots in the given equation to cross check.
Now on substituting $x=4$ in the left-hand side of ${{(x-1)}^{2}}=9$,we get:
${{(4-1)}^{2}}$
On simplifying we get:
$9$, which is the right-hand side therefore this solution is correct
And on substituting $x=-2$ in the left-hand side of ${{(x-1)}^{2}}=9$, we get:
${{(-2-1)}^{2}}$
On simplifying we get-
$9$, which is the right-hand side therefore this solution is also correct.
It is to be remembered that all quadratic equations cannot be simplified easily and not all quadratic equations have real roots. For these types of situations, we use the quadratic formula which is $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.