Question

How do you solve ${{x}^{2}}+2x-7=0$ by completing the square?

Answer 1

Hint: To solve the given equation by using the completing 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 by using an algebraic formula and solve for x.

Complete step-by-step solution:
We have been given an equation ${{x}^{2}}+2x-7=0$.
We have to solve the given equation by completing the square method.
Now, first let us add 7 to the given equation we will get
$\begin{align}
  & \Rightarrow {{x}^{2}}+2x-7+7=0+7 \\
 & \Rightarrow {{x}^{2}}+2x=7 \\
\end{align}$
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. Then we will get
\[\begin{align}
  & \Rightarrow {{\left( \dfrac{b}{2a} \right)}^{2}}={{\left( \dfrac{2}{2} \right)}^{2}} \\
 & \Rightarrow {{\left( 1 \right)}^{2}} \\
 & \Rightarrow 1 \\
\end{align}\]
Now, add 1 to both sides of the equation we will get
$\Rightarrow {{x}^{2}}+2x+1=7+1$
Now, simplifying the above equation we will get
$\Rightarrow {{x}^{2}}+2x+1=8$
Now, we can simplify the LHS to make it the perfect square we will get
$\begin{align}
  & \Rightarrow {{x}^{2}}+2\times \left( 1 \right)x+{{\left( 1 \right)}^{2}}=8 \\
 & \Rightarrow {{\left( x+1 \right)}^{2}}=8 \\
\end{align}$
Now, taking the square root both sides and solving further we will get
\[\begin{align}
  & \Rightarrow x+1=\sqrt{8} \\
 & \Rightarrow x+1=\pm 2\sqrt{2} \\
 & \Rightarrow x=-1\pm 2\sqrt{2} \\
\end{align}\]
So we get two values of x as
$\Rightarrow x=-1+2\sqrt{2},x=-1-2\sqrt{2}$
Hence the solutions of the given equation are $x=-1+2\sqrt{2},-1-2\sqrt{2}$.

Note: The point to remember is that in order to solve the given equation by using the completing the square method the coefficient of ${{x}^{2}}$ must be 1, if not so, then we need to make it 1. When we solve the square root we need to consider both signs $\pm $ in the roots.