Question

Please provide an explanation for the answer! if its right ill reward you 50 points

Answer 1

First note that
`\sqrt x` is only defined for
`x\in\mathbb R` when
`x\ge0`, and that
`\sqrt x\ge0` as a consequence.

This means the equation actually has no real solutions, so if algebraically manipulating the equation does yield a real solution, it must necessarily be extraneous. Let's see what happens.

Squaring both sides, we get


`\left(√(x^2+5x+5)^5)\right)^2=(-1)^2`

`(x^2+5x+5)^5=1`

Take the (real-valued) fifth root of both sides:


`\sqrt[5]{(x^2+5x+5)^5}=\sqrt[5]1`

`x^2+5x+5=1`

Now solve for
`x`.


`x^2+5x+4=0`

`(x+4)(x+1)=0`

`\implies x=-4,x=-1`

But as I pointed out earlier, these must be extraneous. Plug them into the equation to check:


`√(((-4)^2+5(4)+5)^5)=√((16+20+5)^5)=√(41^5)=41^2√(41)\\eq-1`


`√(((-1)^2+5(-1)+5)^5)=√((1-5+5)^5)=√(1^5)=1\\eq-1`

This means the answer is A.