Question

How do you solve ${{x}^{2}}+8x+18=0$ by completing the square?

Answer 1

Hint: To solve the given equation by using completing the square method first we will transform the equation such that the constant term remains alone at the right side. Then we add or subtract a number from the given equation to make the constant term a perfect square. Then factor the left terms and solve for x.

Complete step by step answer:
We have been given an equation ${{x}^{2}}+8x+18=0$.
We have to solve the given equation by completing the square method.
Now, first let us subtract the given equation by 18 we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+8x+18-18=0-18 \\
 & \Rightarrow {{x}^{2}}+8x=-18 \\
\end{align}$
Now, we need to add ${{\left( \dfrac{b}{2a} \right)}^{2}}$ to both sides of the equation to make the constant term a perfect square we get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{8}{2\times 1} \right)}^{2}} \\
 & \Rightarrow {{\left( \dfrac{8}{2} \right)}^{2}} \\
 & \Rightarrow {{\left( 4 \right)}^{2}} \\
 & \Rightarrow 16 \\
\end{align}\]
Now, add 16 to both sides of the equation we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+8x+16=-18+16 \\
 & \Rightarrow {{x}^{2}}+8x+16=-2 \\
\end{align}$
Now, we can simplify the LHS to make it the perfect square we get
$\begin{align}
  & \Rightarrow {{x}^{2}}+4\times 2x+{{4}^{2}}=-2 \\
 & \Rightarrow {{\left( x+4 \right)}^{2}}=-2 \\
\end{align}$

Now, solving further we get
$\Rightarrow x+4=\sqrt{-2}$
As the value inside the square root is negative so the equation has no real solution.

Note: If method is not specified in the question then we can solve the quadratic equation by any one of the methods from completing the square, quadratic formula or by factorization method. To solve the question by using the method asked in the question students must have to know the steps involved. Avoid silly calculation mistakes while solving the equation.