# How do you solve n+sqrt(2n^2-28)=2?

Aug 11, 2015

$n = - 8$

#### Explanation:

Start by isolating the radical term on one side of the equation. To do that, add $- n$ to both sides

$\textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{n}}} + \sqrt{2 {n}^{2} - 28} = 2 - n$

$\sqrt{2 {n}^{2} - 28} = 2 - n$

Now, before you go on, take a moment to write out the conditions that any possible solution must meet.

These conditions will be enforced by the fact that, for real numbers, you can only take the square root of positive numbers. Moreover, the square root of any positive real number is also positive.

This means that you have

• $2 - n \ge 0 \implies n \le 2$
• $2 {n}^{2} - 28 \ge 0 \implies {n}^{2} \ge 14$

The second condition implies $n \in \left(- \infty , - \sqrt{14}\right] \cup \left[\sqrt{14} , + \infty\right)$.

If you combine these two conditions, you will get that any possible solution must satisfy $n \le - \sqrt{14}$, or $n \le - 3.74$.

Next, square both sides of the equation and solve for the roots of the resulting quadratic.

${\left(\sqrt{2 {n}^{2} - 28}\right)}^{2} = {\left(2 - n\right)}^{2}$

$2 {n}^{2} - 28 = 4 - 4 n + {n}^{2}$

Rearrange to get

${n}^{2} + 4 n - 32 = 0$

You can use the quadratic equation to solve for the two roots

${n}_{1 , 2} = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot \left(- 32\right)}}{2 \cdot 1}$

${n}_{1 , 2} = \frac{- 4 \pm \sqrt{144}}{2} = \frac{- 4 \pm 12}{2} = \left\{\begin{matrix}{n}_{1} = \frac{- 4 - 12}{2} = - 8 \\ {n}_{2} = \frac{- 4 + 12}{2} = 4\end{matrix}\right.$

The only solution that satisfies the condition $n \le - \sqrt{14}$ is $n = - 8$, which means that $n = 4$ is an extraneous solution.

You can check the original equation to see that indeed this is the case

$- 8 + \sqrt{2 \cdot {\left(- 8\right)}^{2} - 28} = 2$

$- 8 + \sqrt{100} = 2$

$- 8 + 10 = 2 \textcolor{g r e e n}{\sqrt{}}$

and

$4 + \sqrt{2 \cdot {4}^{2} - 28} = 2$

$4 + 2 \textcolor{red}{\ne} 2$