$
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.

2. 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$

$8x=9$

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

3. 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
$

4. 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
$

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

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

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

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

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

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

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

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

5. 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...

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

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

8. 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 $\frac{x}{1} - \frac{x}{9}$