226k views
5 votes
Please provide an explanation for the answer! if its right ill reward you 50 points

Please provide an explanation for the answer! if its right ill reward you 50 points-example-1
User Roledenez
by
6.8k points

1 Answer

4 votes
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.
User Mark Berry
by
6.0k points