Mathematics
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OpenStudy (seratul):
Help With Algebra
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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?
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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,
\[3\sqrt{x+2} = -\sqrt{x-4}\]
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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
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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):
\[9x+18 = x-4\]
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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?
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OpenStudy (seratul):
it should be -11/4 right?
OpenStudy (3mar):
Correct!
You hit the target 100%
OpenStudy (seratul):
thanks
OpenStudy (3mar):
Welcome!