Thread: 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?

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.

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.

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