Radical Equation Help

**nmbala** · November 18th 2009, 10:48 AM

$ 3\sqrt{x} = \sqrt{x+9} $

Evaluating this i can see that both sides of the equation are equal if we de simplify the left side or simplify the right side since 9 is a perfect square....

But problem is i originally attempted to solve this problem by squaring both sides to get rid of the sqrt leaving me with:

$ 9x = x+9 $

I essentially distributed the exponent to the 3 and the sqrt of x individually. Am I not able to do this must i treat them as one somehow?

This is not the same as we now have an inequality. Can someone explain to me what i did wrong... I know it is simpler to just simplify since its a perfect square, but i am not satisfied doin things one way math should always work no matter which way we try to solve an equation.

**masters** · November 18th 2009, 10:55 AM

Originally Posted by nmbala

$ 3\sqrt{x} = \sqrt{x+9} $

Evaluating this i can see that both sides of the equation are equal if we de simplify the left side or simplify the right side since 9 is a perfect square....

By problem is i originally attempted to solve this problem by squaring both sides to get rid of the sqrt leaving me with:

$ 9x = x+9 $

This is not the same as we now have an inequality. Can someone explain to me what i did wrong... I know it is simpler to just simplify since its a perfect square, but i am not satisfied doin things one way math should always work no matter which way we try to solve an equation.

Hi nmbala,

Where do you have an inequality?

$ 9x=x+9$

$8x=9$

$x=\frac{9}{8}$

**nmbala** · November 18th 2009, 11:06 AM

Hmm i was doin this which must be where i am confused.

$ 9x=x+9 $

$ \frac{9x}{9} = \frac {x+9}{9} $

$ 1x = 1x +9 $

**masters** · November 18th 2009, 11:34 AM

Originally Posted by nmbala

Hmm i was doin this which must be where i am confused.

$ 9x=x+9 $

$ \frac{9x}{9} = \frac {x+9}{9} $

$ 1x = 1x +9 $

Your last step does not follow from the previous one.

$\frac{9x}{9}=\frac{x+9}{9}$

$1x=\frac{x+9}{9}$

$x=\frac{x}{9}+1$

$x-\frac{x}{9}=1$

$\frac{8}{9}x=1$

$x=\frac{1}{\frac{8}{9}}$

$x=\frac{9}{8}$

But I would never go about it that way......too long and unnecessary.

**nmbala** · November 18th 2009, 11:39 AM

Opps was suppose to be +1 not 9... Regardless i see that i need to carry that 9 to the x also. Thanks for your help!

Last question regarding steps 4 and 5 of your previous post. What is the logic behind:
$x-\frac{x}{9}=1$

$\frac {8}{9}x =1$

Appears I have forgotten something else from class 3 years ago...

**Raoh** · November 18th 2009, 11:54 AM

Hello

$x-\frac{x}{9}=1\Leftrightarrow \frac{9x-x}{9}=1\Leftrightarrow \frac{8x}{9}=1$

**Defunkt** · November 18th 2009, 11:55 AM

$x - \frac{x}{9} = \frac{9}{9}x - \frac{1}{9}x = \left( \frac{9}{9} - \frac{1}{9} \right)x = \frac{8}{9}x$

**nmbala** · November 18th 2009, 11:56 AM

Originally Posted by Raoh

Hello

$x-\frac{x}{9}=1\Leftrightarrow \frac{9x-x}{9}=1\Leftrightarrow \frac{8x}{9}=1$

Right Lowest common denominator to add unlike fractions. Thanks!

I need to start remembering to look at things as $\frac{x}{1} - \frac{x}{9}$

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