Question

How do you solve ${{x}^{2}}-2x-14=0$ by completing the square?

Answer 1

Hint: We keep the variables and the constants separate on both sides of the equality. We then add 1 to both sides of the equation ${{x}^{2}}-2x=14$. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we add 1 to the both sides to find the value of $x$ for ${{x}^{2}}-2x=14$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}-2x-14=0$.
Separating the variables and the constants we get ${{x}^{2}}-2x=14$.
We first add 1 to both sides of ${{x}^{2}}-2x=14$. We use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$.
$\begin{align}
  & {{x}^{2}}-2x+1=14+1 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}=15 \\
\end{align}$
We interchanged the numbers for $a=x,b=1$.
Now we have a quadratic equation ${{\left( x-1 \right)}^{2}}=15$.
We need to find the solution of the given equation ${{\left( x-1 \right)}^{2}}=15$.
We take square root on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x-1 \right)}^{2}}}=\pm \sqrt{15} \\
 & \Rightarrow \left( x-1 \right)=\pm \sqrt{15} \\
\end{align}$
Now we add 1 to the both sides of the equation $\left( x-1 \right)=\pm \sqrt{15}$ to get value for variable $x$.
$\begin{align}
  & \left( x-1 \right)+1=\pm \sqrt{15}+1 \\
 & \Rightarrow x=1\pm \sqrt{15} \\
\end{align}$

The given quadratic equation has two solutions and they are $x=1\pm \sqrt{15}$.

Note: We can assume the function of $f\left( x \right)={{x}^{2}}-2x-14$. Now we can validate the solution of $x=1\pm \sqrt{15}$ for $f\left( x \right)={{x}^{2}}-2x-14$. We put the value of $x=1+\sqrt{15}$ in the equation of $f\left( x \right)={{x}^{2}}-2x-14$.
Now we find the value of \[f\left( 1+\sqrt{15} \right)={{\left( 1+\sqrt{15} \right)}^{2}}-2\left( 1+\sqrt{15} \right)-14\]
\[\begin{align}
  & {{\left( 1+\sqrt{15} \right)}^{2}}-2\left( 1+\sqrt{15} \right)-14 \\
 & =1+15+2\sqrt{15}-2-2\sqrt{15}-14 \\
 & =0 \\
\end{align}\]
Therefore, the root value $x=1+\sqrt{15}$ satisfies $f\left( x \right)={{x}^{2}}-2x-14$. Same thing can be said about $x=1-\sqrt{15}$.