Question

How do you solve ${{x}^{2}}=14x+5$?

Answer 1

Hint: We keep the variables and the constants separate on both sides of the equality. Then we add 49 to both sides of the equation ${{x}^{2}}-14x=5$. 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 7 to the both sides to find the value of $x$ for ${{x}^{2}}=14x+5$.

Complete step by step solution:
We need to find the solution of the given equation ${{x}^{2}}=14x+5$.
We separate the variables and the constants on both sides of the equality to get ${{x}^{2}}-14x=5$.
We first add 49 to both sides of ${{x}^{2}}-14x=5$. We use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$.
$\begin{align}
  & {{x}^{2}}-14x+49=5+49 \\
 & \Rightarrow {{x}^{2}}-2\times 7\times x+{{7}^{2}}=54 \\
 & \Rightarrow {{\left( x-7 \right)}^{2}}=54 \\
\end{align}$
We interchanged the numbers for $a=x,b=7$.
Now we have a quadratic equation ${{\left( x-7 \right)}^{2}}=54$.
We need to find the solution of the given equation ${{\left( x-7 \right)}^{2}}=54$.
We take square roots 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.
$\begin{align}
  & \sqrt{{{\left( x-7 \right)}^{2}}}=\pm \sqrt{54} \\
 & \Rightarrow \left( x-7 \right)=\pm 3\sqrt{6} \\
\end{align}$
Now we add 7 to the both sides of the equation $\left( x-7 \right)=\pm 3\sqrt{6}$ to get value for variable $x$.
$\begin{align}
  & \left( x-7 \right)+7=\pm 3\sqrt{6}+7 \\
 & \Rightarrow x=7\pm 3\sqrt{6} \\
\end{align}$

The given quadratic equation has two solutions and they are $x=7\pm 3\sqrt{6}$.

Note: We can assume the function of ${{x}^{2}}=14x+5$. Now we can validate the solution of $x=7\pm 3\sqrt{6}$ for $f\left( x \right)={{x}^{2}}-14x-5$. We put the value of $x=7+3\sqrt{6}$ in the equation of $f\left( x \right)={{x}^{2}}-14x-5$.
Now we find the value of \[f\left( 7+3\sqrt{6} \right)={{\left( 7+3\sqrt{6} \right)}^{2}}-14\left( 7+3\sqrt{6} \right)-5\]
\[\begin{align}
  & {{\left( 7+3\sqrt{6} \right)}^{2}}-14\left( 7+3\sqrt{6} \right)-5 \\
 & =49+54+42\sqrt{6}-98-42\sqrt{6}-5 \\
 & =0 \\
\end{align}\]
Therefore, the root value $x=7+3\sqrt{6}$ satisfies $f\left( x \right)={{x}^{2}}-14x-5$. Same thing can be said about $x=7-3\sqrt{6}$.