Question

Solve the equation identify extraneous solutions

Answer 1

Answer:

Step-by-step explanation:

The goal to solving any equation is to have x = {something}.  That means we need to get the x out from underneath that radical.  It's a square root, so we can "undo" it by squaring.  Square both sides because this is an equation.  Squaring both sides gives you

$x^2=-3x+40$

Get everything on one side of the equals sign and set the quadratic equal to 0:

$x^2+3x-40=0$

Throw this into the quadratic formula to get that the solutions are x = 5 and -8.  We need to see if only one works, both work, or neither work in the original equation.

Does $5=\sqrt{-3(5)+40}$?

$5=\sqrt{-15+40}$ and

$5=\sqrt{25}$

and 5 = 5.  So 5 works.  Let's try -8 now:

$-8=\sqrt{-3(-8)+40}$ and

$-8=\sqrt{24+40}$ so

$-8=\sqrt{64}$

-8 = 8?  No it doesn't.  So only 5 works. Your choice is the third one down.