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

##### 1 Answer

#### Answer:

#### Explanation:

Start by isolating the radical term on one side of the equation. To do that, add

#color(red)(cancel(color(black)(n))) - color(red)(cancel(color(black)(n))) + sqrt(2n^2-28) = 2-n#

#sqrt(2n^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>=0 implies n <=2# #2n^2-28>=0 implies n^2>=14#

The second condition implies

If you combine these two conditions, you will get that any possible solution must satisfy

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

#(sqrt(2n^2 - 28))^2 = (2-n)^2#

#2n^2 - 28 = 4 - 4n + n^2#

Rearrange to get

#n^2 + 4n - 32 = 0#

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

#n_(1,2) = (-4 +- sqrt(4^2 - 4 * 1 * (-32)))/(2 * 1)#

#n_(1,2) = (-4 +- sqrt(144))/2 = (-4 +- 12)/2 = {(n_1 = (-4 - 12)/2 = -8), (n_2 = (-4 + 12)/2 = 4) :}#

The only solution that satisfies the condition *extraneous solution*.

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

#-8 + sqrt(2 * (-8)^2 - 28) = 2#

#-8 + sqrt(100) = 2#

#-8 + 10 = 2 color(green)(sqrt())#

and

#4 + sqrt(2 * 4^2 - 28) = 2#

#4 + 2 color(red)(!=) 2#