Question

How do you solve `sqrt(5x^2+11)=x+5` and identify any restrictions?

Answer 1

`x=-1` and `x=7/2`. This equation has no restrictions since `5x^2+11>=0`.

Explanation:

Begin by squaring both sides of the equation to eliminate the square root. Recall that in squaring the right hand side, we can't square each term individually--we have to square the entire right hand side.

`(sqrt(5x^2+11))^2=(x+5)^2`

On the left, the `sqrt` and `""^2` cancel. On the right, expand the square by FOILing:

`5x^2+11=(x+5)(x+5)=x^2+5x+5x+25`

`5x^2+11=x^2+10x+25`

Putting all the terms on the same side:

`4x^2-10x-14=0`

Divide all terms by `2`:

`2x^2-5x-7=0`

Which we can factor by splitting the middle term:

`2x^2+2x-7x-7=0`

`2x(x+1)-7(x+1)=0`

`(2x-7)(x+1)=0`

Which give `x=7/2` and `x=-1`.

Check both of these by plugging them into the original equation:

Checking `x=7/2`:

`sqrt(5(7/2)^2+11)=7/2+5`

`sqrt(5(49/4)+11)=7/2+10/2`

`sqrt(245/4+44/4)=7/2+10/2`

`sqrt(289/4)=17/2`

Which is true! So `x=7/2` is a valid solution.

Checking `x=-1`:

`sqrt(5(-1)^2+11)=-1+5`

`sqrt(5+11)=4`

`sqrt16=4`

Which is true as well, so our solutions are `x=-1` and `x=7/2`.

We can skip the process of going back and checking answers by noting that since we have `sqrt(5x^2+11)` in the original equation, we can't have values of `x` where `5x^2+11<0`, since we can't take the square root of a negative value.

Note that since `5x^2` and `11` are both always positive, `5x^2+11` will also always be positive. Thus, there is never a time when it's negative and by the same coin there is never a time when `sqrt(5x^2+11)` won't be defined. Thus there are no restrictions on this equation.