# How do you solve #sqrt(2x+2)= sqrt( x^2-6)#?

##### 1 Answer

#### Explanation:

Since you're dealing with *square roots*, it's always a good idea to start by writing down the **valid** solution intervals for

When working with *real numbers*, you can only take the square root of **positive numbers**, which means that you'll need

#2x + 2 >=0#

#2x >= -2 implies x >= -1#

and

#x^2 - 6 >=0#

#(x-sqrt(6))(x+sqrt(6)) >=0#

This takes place when you have

#x in (-oo, -sqrt(6)] uu [sqrt(6), + oo)#

Combine these two restrictions,

#color(purple)(|bar(ul(color(white)(a/a)color(black)(x in [sqrt(6), + oo))color(white)(a/a)|)))#

In order for a value of **valid solution** to the original equation, you need it to come from that interval.

Now, square both sides of the equation to get rid of the square roots

#(sqrt(2x+2))^2 = (sqrt(x^2 - 6))^2#

#2x + 2 = x^2 - 6#

Rearrange to quadratic equation form

#x^2 - 2x - 8 = 0#

You can calculate the two solutions by using the **quadratic formula**

#color(blue)(|bar(ul(color(white)(a/a)x_(1,2) = (-b +- sqrt(b^2 - 4 * a * c))/(2 * a)color(white)(a/a)|)))#

Here

In your case, you'll have

#x_(1,2) = (-(-2) +- sqrt( (-2)^2 - 4 * 2 * (-8)))/(2 * 1)#

#x_(1,2) = (2 +- sqrt(36))/2#

#x_(1,2) = (2 +- 6)/2 implies { (x_1 = (2 - 6)/2 = color(red)(cancel(color(black)(-2)))), (x_2 = (2 + 6)/2 = 4color(white)(a)color(green)(sqrt())) :}#

Now, only one of these two values will be a valid solution to the original equation. Since

#x = -2" " !in [sqrt(6), + oo)#

you can say that *extraneous solution*. The only **valid solution** to the original equation will thus be

#color(green)(|bar(ul(color(white)(a/a)color(black)(x = 4)color(white)(a/a)|)))#

Do a quick double-check to make sure that the calculations are correct

#sqrt(2 * 4 + 2) = sqrt(4^2 - 6)#

#sqrt(10) = sqrt(10) color(white)(a)color(green)(sqrt())#