I have a problem that says

√x+8) - √x-4) = 2

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

Thank you in advance.

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- Sep 12th 2010, 11:45 AMJoacoSquare roots subtraction
I have a problem that says

√x+8) - √x-4) = 2

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

Thank you in advance. - Sep 12th 2010, 11:52 AMe^(i*pi)
Do you mean

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

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

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

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

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

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

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

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 PMHallsofIvy
I would have done that slightly differently. From $\displaystyle \sqrt{x+ 8}- \sqrt{x- 4}= 2$, add $\displaystyle \sqrt{x- 4}$ to both sides to get $\displaystyle \sqrt{x+ 8}= \sqrt{x- 4}+ 2$. Squaring both sides of that you get $\displaystyle x+ 8= x- 4+ 4\sqrt{x- 4}+ 4$. 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 $\displaystyle 8= 4\sqrt{x- 4}$ which reduces to $\displaystyle \sqrt{x- 4}= 2$. Now square both sides again to get x- 4= 4.

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