# is this a way to solve a radical equation?

• Apr 15th 2011, 08:25 PM
bournouli
is this a way to solve a radical equation?
(2x^2-7)^(1/2)=3-x

so FOIL the left side, square the right side for, 2x^2-7=x^2-6x+9. Add 7: 2x^2=x^2-6x+16 but then divide by 2 instead of adding 2x^2: x^2=(x^2-6x+16)(1/2).
Then subtract x^2 and get .5x^2-3x+8=0. Then use the quadratic formula to solve for x.

Is this a licit way to solve this? If not, then why not?
• Apr 15th 2011, 08:29 PM
Prove It
No. You can't FOIL the left hand side because it's not the product of two binomials...

What you actually have is Sqrt(2x^2 - 7) = 3 - x.

Start by squaring both sides to undo the square root, then expand the RHS and solve the resulting quadratic.
• Apr 16th 2011, 08:10 AM
bournouli
Quote:

Originally Posted by Prove It
No. You can't FOIL the left hand side because it's not the product of two binomials...

What you actually have is Sqrt(2x^2 - 7) = 3 - x.

Start by squaring both sides to undo the square root, then expand the RHS and solve the resulting quadratic.

Don't you have to foil the left side then, since you are squaring both sides? That's what I meant in any case.
• Apr 16th 2011, 08:14 AM
topsquark
Quote:

Originally Posted by bournouli
Don't you have to foil the left side then, since you are squaring both sides? That's what I meant in any case.

You do not FOIL the LHS. That is what Prove It is trying to say.
sqrt(2x^2 - 7) = 3 - x

[ sqrt(2x^2 - 7) ]^2 = (3 - x)^2

2x^2 - 7 = (3 - x)^2

Now FOIL the RHS... Also, when you are done check your solutions to make sure they work. Squaring both sides of an equation often introduces extra solutions.

-Dan