I am stuck on this problem as I think there is no solution can you please tell me if I am right.

Solve for x:√x-6 - √x-6 = 36

and problem # 2

Solve for x √x-6√x-6 = 15

Please help me.

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- Jul 2nd 2008, 02:14 PMtricey36radical equations
I am stuck on this problem as I think there is no solution can you please tell me if I am right.

Solve for x:√x-6 - √x-6 = 36

and problem # 2

Solve for x √x-6√x-6 = 15

Please help me. - Jul 2nd 2008, 02:17 PMMathstud28
- Jul 2nd 2008, 02:53 PMticbol
- Jul 2nd 2008, 03:22 PMReckoner
As written, these have no solutions (which the others have pointed out):

$\displaystyle \sqrt{x} - 6 - \sqrt x - 6 = 36\Rightarrow 0 = 36$

For the second, assuming you wrote it correctly,

$\displaystyle \sqrt{x} - 6\sqrt x - 6 = 15\Rightarrow -5\sqrt{x} - 6 = 15$

$\displaystyle \Rightarrow\sqrt x = -\frac{21}5$

which has no real solutions. - Jul 2nd 2008, 05:12 PMtopsquark
In the future, PLEASE use parenthesis!

Let me try to read the tea leaves. tricey36, please confirm which of these (if any) you meant.

First problem:

$\displaystyle \sqrt{x} - 6 - \sqrt{x - 6} = 36$

or

$\displaystyle \sqrt{x - 6} - \sqrt{x - 6} = 36$

(Though this one's just silly.)

or

$\displaystyle \sqrt{x - 6} - \sqrt{x} - 6 = 36$

Second problem:

$\displaystyle \sqrt{x - 6} \sqrt{x - 6} = 15$

or

$\displaystyle \sqrt{x} - 6\sqrt{x} - 6 = 15$

These are all I can (reasonably) think of.

-Dan