Question

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

Answer 1

`x = 4`

Explanation:

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

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, `x >=-1` and `x in (-oo, -sqrt(6)] uu [sqrt(6), + oo)` to get

`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 `x` to be a 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 `a`, `b`, an `c` are the coefficients of the quadratic equation `ax^2 + bx + c = 0`.

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 `x = -2` will be an 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())`