Question

How do you solve `sqrt(x+4 )= 1- sqrt(3x+13)`?

Answer 1

`x=-4`

Explanation:

Given:

`sqrt(x+4) = 1-sqrt(3x+13)`

We can square both sides to get:

`x+4=1-2sqrt(3x+13)+(3x+13) = 3x+14-2sqrt(3x+13)`

Add `2sqrt(3x+13)-x-4` to both ends to get:

`2sqrt(3x+13) = 2x+10`

Divide both sides by `2` to get:

`sqrt(3x+13) = x+5`

Square both sides to get:

`3x+13 = x^2+10x+25`

Subtract `3x+13` from both sides to get:

`0 = x^2+7x+12 = (x+3)(x+4)`

So:

`x = -3" "` or `x = -4`

We now need to check for extraneous roots, since these may have been introduced when we squared the equation:

`sqrt((color(blue)(-3))+4) = 1 != -1 = 1-2 = 1-sqrt(4) = 1-sqrt(3(color(blue)(-3))+4)`

`sqrt((color(blue)(-4))+4) = 0 = 1-1 = 1-sqrt(1) = 1-sqrt(3(color(blue)(-4))+13)`

So `x=-3` is an extraneous solution caused by squaring `1 != -1`, while `x=-4` is a solution of the original equation.