how do i solve... the square root of 3x - 5= x - 5

Question

how do i solve...  the square root of 3x - 5= x - 5

lgbasallote · Answer

$$\sqrt{3x - 5} = x -5$$
???

barrycarter · Answer

Assuming @lgbasallote said it correctly, start by squaring both sides

anonymous · Answer

yes... how did you get the square root sign

lgbasallote · Answer

by following your instructions?

anonymous · Answer

ok whatever...

anonymous · Answer

ok i squared both sides

anonymous · Answer

hello??

barrycarter · Answer

What did you get after squaring?

anonymous · Answer

3x-5=(x+-5)(x+-5

barrycarter · Answer

OK, and when you multiply it out, that's a quadratic equation, right?

anonymous · Answer

yeah i think... i got x^2+-13x+30... is that right?

barrycarter · Answer

Yes, it does. (you can always check final answers at wolframalpha.com)

anonymous · Answer

so is that all i do or do i solve for x...???

barrycarter · Answer

You have to solve by x either by factoring, completing the square, or applying the quadratic formula

anonymous · Answer

i am trying to make it to where someone could do foil...but im having difficulties factoring it in...!

barrycarter · Answer

What two numbers multiply out to 30, but add to -13 (hint: -3 is one of the numbers)

anonymous · Answer

o is it -3 and -10

anonymous · Answer

so it would be (x-3)(x-10)?????

barrycarter · Answer

Correct, (x-3)*(x-10) = 0

anonymous · Answer

so is that all my answer is when it tells me to solve.

gw2011 · Answer

x^2-13x+30 = 0 factoring the left side you get:
(x-10 (x-3) = 0
x-10 = 0
x = 10

x-3 = 0
x = 3

When you substitute each of these values,x=10 and x=3, into the original equation, then the only one that has the right solution is; x=10

For x=10 you get from the original equation 5=5
But for x=3 you get from the original equation 2=-2 which is not correct

anonymous · Answer

o ok thank you everyone!!!! :)

barrycarter · Answer

Good catch, @gw2011 -- I forgot about extraneous roots for a moment