Question

How do you solve `3+sqrt[x+7]=sqrt[x+4]` and find any extraneous solutions?

Answer 1

the equation is impossible

Explanation:

you can calculate

`(3+sqrt(x+7))^2=(sqrt(x+4))^2`

`9+x+7+6sqrt(x+7)=x+4`

that's

`6sqrt(x+7)=cancel(x)+4-9cancel(-x)-7`

`6sqrt(x+7)=-12`

that's impossible because a square root must be positive

Answer 2

No real roots of `x` exist in `R` (`x!inR`)

`x` is a complex number `x=4*i^4-7`

Explanation:

First to solve this equation we think how to take off the square root, by squaring both sides:

`(3+sqrt(x+7))^2=(sqrt(x+4))^2`

Using the binomial property for squaring of sum
`(a+b)^2=a^2+2ab+b^2`

Applying it on both sides of the equation we have:

`(3^2+2*3*sqrt(x+7)+(sqrt(x+7))^2)=x+4`

Knowing that `(sqrt(a))^2=a`
`9+6sqrt(x+7)+x+7=x+4`

Taking all the know a and unknowns to the second side leaving the square root on one side we have:

`6sqrt(x+7)=x+4-x-7-9`
`6sqrt(x+7)=-12`
`sqrt(x+7)=-12/6`
`sqrt(x+7)=-2`

Since square root equal to a negative real number that is
impossible in `R`, no roots exists so we have to check complex set.

`sqrt(x+7)=-2`

Knowing that i^2=-1 that means `-2=2*i^2`

`sqrt(x+7)=2i^2`
Squaring both sides we have:

`x+7=4*i^4`
Therefore , `x=4*i^4-7`

So `x` is a complex number.