Question

How do you solve ${{x}^{2}}-6x+11=0$ by completing the square?

Answer 1

Hint: To solve the equation using a completing square, the first step is to take the constant term to the other side of the equation, the second step is taking half of the value of the coefficient of x and square it. Then add the resulting term to both sides of the equation. Simplify the quadratic equation by algebraic identity and take square root on both sides. Finally, get the value of x by moving the constant term on the other side of the equation.

Complete step by step answer:
According to the question we need to use completing the square for the equation ${{x}^{2}}-6x+11=0$. Take 11 to the other side of the equation so the sign of that number will be changed to negative which is${{x}^{2}}-6x=-11$. Now, take half of the value of x or just divide the coefficient of x by 2 we get $\dfrac{-6}{2}=-3$ .
Squaring and adding -3 to each side of the equation,
${{x}^{2}}-6x+{{\left( -3 \right)}^{2}}=-11+{{\left( -3 \right)}^{2}}$
While squaring the numbers negative becomes positive.
$\Rightarrow {{x}^{2}}-6x+{{\left( 3 \right)}^{2}}=-11+{{\left( 3 \right)}^{2}}$
We know the algebraic identity,
${{(a+b)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Using algebraic identity on the left-hand side of the equation and substitute the x in place of a and 3 in the place of b then we get the equation in the form of ${{x}^{2}}-6x+{{\left( 3 \right)}^{2}}={{\left( x-3 \right)}^{2}}$.
$\Rightarrow {{\left( x-3 \right)}^{2}}=-11+{{\left( 3 \right)}^{2}}$
On further simplifications, we get
$\Rightarrow {{\left( x-3 \right)}^{2}}=-11+9$
On solving 11 and 9 we get this
$\Rightarrow {{\left( x-3 \right)}^{2}}=-2$
Now, taking square root on both sides of the equation, \[\]
$\Rightarrow \sqrt{{{\left( x-3 \right)}^{2}}}=\pm \sqrt{-2}$
Now on the left-hand side square and the square root is vanished. So, the equation will be
$\Rightarrow \left( x-3 \right)=\pm \sqrt{-2}$
$\Rightarrow \left( x-3 \right)=\pm \sqrt{2}\times \sqrt{-1}$
And we know that $\sqrt{-1}=i$
$\Rightarrow \left( x-3 \right)=\pm \sqrt{2}\times i$
Now, taking the constant term to the right-hand side of the equation then the sign of that term will change,
$\therefore x=\pm i\sqrt{2}+3$

Hence, the required value is $x=\pm i\sqrt{2}+3$ for ${{x}^{2}}-6x+11=0$.

Note: In completing squares the coefficient ${{x}^{2}}$ should be 1. Otherwise, divide that number with the whole equation then proceed with the problem. Instead of taking the constant term to the other side of the equation, we can add that constant term on both sides of the equation in the first step. We can verify for our answer by putting the obtained value of x in the given quadratic equation.