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Mathematics 12 Online
OpenStudy (seratul):

Help With Algebra

OpenStudy (seratul):

\[3\sqrt{x+2} + \sqrt{x-4} = 0\]

OpenStudy (3mar):

If I may help?

OpenStudy (seratul):

Yes pleaseeeeeeeee

OpenStudy (3mar):

Thank you.

OpenStudy (3mar):

Let's begin Of course you need to solve for x, right?

OpenStudy (seratul):

yea

OpenStudy (3mar):

Ok can you send the second square root to the other side?

OpenStudy (seratul):

\[3\sqrt{x+2} = \sqrt{x-4}\]

OpenStudy (3mar):

with a negative sign, not as it is!

OpenStudy (seratul):

my baaad, \[3\sqrt{x+2} = -\sqrt{x-4}\]

OpenStudy (3mar):

Great!

OpenStudy (3mar):

Now can you raise the two side for the power of two?

OpenStudy (seratul):

\[3(x+2) = -(x-4)\]

OpenStudy (seratul):

Helloo?

OpenStudy (seratul):

@Jamierox4ev3r @Directrix

OpenStudy (3mar):

Sorry for late. It was an emergency call.

Directrix (directrix):

Is the left side of the equation the cube root of (x + 2) or is it 3 times square root of (x + 2) ?

OpenStudy (3mar):

You ought to square the 3 also, not just the root!

OpenStudy (seratul):

I wrote the equation exactly like the problem. It was next to it, so I'm pretty sure it's times.

OpenStudy (seratul):

\[9x+18 = x-4\]

OpenStudy (seratul):

is that right?

OpenStudy (3mar):

Correct! and the final step .. rearrange and collect x in one side

OpenStudy (seratul):

tyyyy!! I can do the rest.

OpenStudy (3mar):

You are welcome! Any Help... Any Time...

OpenStudy (3mar):

Any more questions?

OpenStudy (seratul):

it should be -11/4 right?

OpenStudy (3mar):

Correct! You hit the target 100%

OpenStudy (seratul):

thanks

OpenStudy (3mar):

Welcome!

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