# Square roots subtraction

• Sep 12th 2010, 11:45 AM
Joaco
Square roots subtraction
I have a problem that says
√x+8) - √x-4) = 2

I don't know how to continue with this problem...

• Sep 12th 2010, 11:52 AM
e^(i*pi)
Quote:

Originally Posted by Joaco
I have a problem that says
√x+8) - √x-4) = 2

I don't know how to continue with this problem...

Do you mean

A. $\sqrt{x+8} - \sqrt{x-4} = 2$ or

B. $\sqrt{x}+8 - (\sqrt{x}-4) = 2$ or

C. $\sqrt{x}+8-\sqrt{x}-4 = 2$

If, as I suspect, the correct expression is A then you will need to square both sides

$(x+8) - 2\sqrt{(x+8)(x-4)} + (x-4) = 4$

Simplfying: $\sqrt{(x+8)(x-4)} = x$

Square again: $(x+8)(x-4) = x^2$

I'll leave you to finish that off

edit: this equation has only one solution and it is in the domain of the original equation so there can be no extraneous solutions
• Sep 12th 2010, 12:45 PM
HallsofIvy
I would have done that slightly differently. From $\sqrt{x+ 8}- \sqrt{x- 4}= 2$, add $\sqrt{x- 4}$ to both sides to get $\sqrt{x+ 8}= \sqrt{x- 4}+ 2$. Squaring both sides of that you get $x+ 8= x- 4+ 4\sqrt{x- 4}+ 4$. You still have the square root form the "cross term" but since you don't have both square roots on the same side, it's a little simpler. That can be reduced, by canceling the "x" terms on both sides and canceling the "-4" and "4", to $8= 4\sqrt{x- 4}$ which reduces to $\sqrt{x- 4}= 2$. Now square both sides again to get x- 4= 4.

Notice that while e^(i*pi)'s equation $(x+ 8)(x- 4)= x^2$ looks like a quadratic, it is not be cause the $x^2$ terms on each side will cancel.
• Sep 12th 2010, 01:22 PM
Joaco
Quote:

Originally Posted by HallsofIvy
I would have done that slightly differently. From $\sqrt{x+ 8}- \sqrt{x- 4}= 2$, add $\sqrt{x- 4}$ to both sides to get $\sqrt{x+ 8}= \sqrt{x- 4}+ 2$. Squaring both sides of that you get $x+ 8= x- 4+ 4\sqrt{x- 4}+ 4$. You still have the square root form the "cross term" but since you don't have both square roots on the same side, it's a little simpler. That can be reduced, by canceling the "x" terms on both sides and canceling the "-4" and "4", to $8= 4\sqrt{x- 4}$ which reduces to $\sqrt{x- 4}= 2$. Now square both sides again to get x- 4= 4.

Notice that while e^(i*pi)'s equation $(x+ 8)(x- 4)= x^2$ looks like a quadratic, it is not be cause the $x^2$ terms on each side will cancel.

That helped me out a lot, I was stuck at the $8= 4\sqrt{x- 4}$ part until I read your post.
Thank you a lot.

Also thank you a lot $\sqrt{x+ 8}- \sqrt{x- 4}= 2$.