SOLUTION: I am stuck on this problem. I did the equation and my answer was y=0,y=9. The problem is to use factoring to solve the equation. 2y^2-162=0 If I'm correct, yea for me!

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Question 153542This question is from textbook Intermediate Algebra
: I am stuck on this problem. I did the equation and my answer was y=0,y=9. The problem is to use factoring to solve the equation. 2y^2-162=0
If I'm correct, yea for me! This question is from textbook Intermediate Algebra

Found 3 solutions by Fombitz, stanbon, josmiceli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Easiest way is to plug your answer in the original equation.
2y%5E2-162=0
2%280%29%5E2-162=0
-162=0
Untrue.
y=0 is not a solution.
.
.
.
2y%5E2-162=0
2%289%29%5E2-162=0
2%2881%29-162=0
162-162=0
0=0
True.
y=9 is a solution.
.
.
.
You're missing an answer.
2y%5E2-162=0
2y%5E2=162
y%5E2=81
y=0+%2B-+9
You forgot (-9).
.
.
.
Here's a graph of the function.
+graph%28+300%2C+300%2C+-10%2C+10%2C+-180%2C+180%2C+2x%5E2-162%29+

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Also y = -9.
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
2y%5E2+-+162+=+0
I know how to solve it, but I'm not sure what the
"factoring" is about. I guess it means to factor out a 2
from the left side like this:
2%28y%5E2+-+81%29+=+0
Now if I divide both sides by 2, I get
y%5E2+-+81+=+0 (since 0%2F2+=+0)
y%5E2+=+81
Take the square root of both sides
y+=+9
and
y+=+-9
I think you were half right