Radicals

**OzzMan** · February 2nd 2008, 04:57 PM

solve the given equation for x
(x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2)
Sorry I didnt write it in radical form, wasnt sure how. btw the -(x)^(1/2) is in the numerator. the answer in the back of the book is 25. Half way thru i got x + (x)^(1/2) - 45 = 0 and then substituted y = (x)^(1/2) and y^2=x for the values. But my end results didnt seem to be right. Sorry if this is cluttered.

**mr fantastic** · February 2nd 2008, 05:26 PM

Originally Posted by OzzMan

solve the given equation for x
(x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2)
Sorry I didnt write it in radical form, wasnt sure how. btw the -(x)^(1/2) is in the numerator. the answer in the back of the book is 25. Half way thru i got x + (x)^(1/2) - 45 = 0 and then substituted y = (x)^(1/2) and y^2=x for the values. But my end results didnt seem to be right. Sorry if this is cluttered.

First, if the answer is x = 25, then the equation is NOT (x-9)^(1/2)=36/(x-9)^(1/2) - (x)^(1/2), it's (x-9)^(1/2)=36/(x-9)^(1/2) + (x)^(1/2).

$\sqrt{x-9} = \frac{36}{\sqrt{x-9} + \sqrt{x}}$

$\Rightarrow \sqrt{x-9} (\sqrt{x-9} + \sqrt{x})= 36$

$\Rightarrow x - 9 + \sqrt{x-9} \sqrt{x} = 36$

$\Rightarrow \sqrt{x-9} \sqrt{x} = 45 - x$

$\Rightarrow (x - 9)x = (45 - x)^2$

$\Rightarrow x^2 - 9x = 45^2 - 90x + x^2$

$\Rightarrow -9x = 45^2 - 90x$

$\Rightarrow 81x = 45^2 \Rightarrow 9^2 x = 45^2 \Rightarrow x = \left( \frac{45}{9} \right)^2 = 25$ .

**OzzMan** · February 2nd 2008, 05:33 PM

you wrote it wrong you have + radical x in the denominator its - radical x and its not part of that fraction its over 1

**mr fantastic** · February 2nd 2008, 05:53 PM

Originally Posted by OzzMan

you wrote it wrong you have + radical x in the denominator its - radical x and its not part of that fraction its over 1

Oh I see. The numerator business in you original post and the its over 1 business in your next post are just red herrings. The equation really is as you originally stated:

$\sqrt{x-9} = \frac{36}{\sqrt{x-9}} - \sqrt{x}$

$\Rightarrow x - 9 = 36 - \sqrt{x-9} \sqrt{x}$

from which it follows that:

Originally Posted by mr fantastic

[snip]

$\Rightarrow \sqrt{x-9} \sqrt{x} = 45 - x$

$\Rightarrow (x - 9)x = (45 - x)^2$

$\Rightarrow x^2 - 9x = 45^2 - 90x + x^2$

$\Rightarrow -9x = 45^2 - 90x$

$\Rightarrow 81x = 45^2 \Rightarrow 9^2 x = 45^2 \Rightarrow x = \left( \frac{45}{9} \right)^2 = 25$ .

**OzzMan** · February 2nd 2008, 05:57 PM

yeah thats the right equation. Again I'm sorry i didn't use latex. I'm just no good at it. I always get an error.

**OzzMan** · February 2nd 2008, 06:05 PM

im a little confused this is probably where i messed up, but i did the cross multiply of radical x-9 and the radicals crossed out like you have but why would radical x-9 stay on the right side also?

**mr fantastic** · February 2nd 2008, 06:13 PM

Originally Posted by OzzMan

im a little confused this is probably where i messed up, but i did the cross multiply of radical x-9 and the radicals crossed out like you have but why would radical x-9 stay on the right side also?

You have to multiply both of the terms on the right hand side by $\sqrt{x - 9}$ ..... So you get a cancelling that leaves you with the 36, AND the $\sqrt{x}$ gets multiplied by $\sqrt{x - 9}$ to leave you with $\sqrt{x - 9} \sqrt{x}$ .

**OzzMan** · February 2nd 2008, 06:22 PM

yea i thought as much. just didnt see why you had to multipy the radical x by radical x-9. didnt think that applied to cross multiplying. but i see where your coming from.

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