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Mathematics
OpenStudy (anonymous):

Solving Square Root Equations 5(squareroot of)x-1=(squaredrootof) x+1

OpenStudy (dumbcow):

square both sides --> 25(x-1) = x+1

OpenStudy (anonymous):

yes, but i need to get x. :(

OpenStudy (dumbcow):

you can solve it from there get x by itself

OpenStudy (anonymous):

it would be 25x? thats somethign i dont get.

OpenStudy (anonymous):

\[5\sqrt{x-1}=\sqrt{x+1}\]\[(5\sqrt{x-1})^{2}=(\sqrt{x+1})^2\]\[25(x-1)=x+1\]\[25x-25=x+1\]-x +25 -x +25\[24x=26\]/24 /24\[x=\]

OpenStudy (anonymous):

i dont get the 5th step...

OpenStudy (anonymous):

The "-x +25 -x +25" step?

OpenStudy (anonymous):

wouldnt it be 24x-24=0? yeah yhat step.

OpenStudy (anonymous):

No\[-25-1=-26\]

OpenStudy (anonymous):

wooow... feeling like a genius...

OpenStudy (anonymous):

If you moved both the x and 1 from the right side to the left you'd subtract -1 from a negative 25, making you go more negative.

OpenStudy (anonymous):

I hate making those simple mistakes.

OpenStudy (anonymous):

okay. so, if i also take the x out, because i want to have =0, it owuld be 24x-26=o, correct?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

so now i factor it, right?

OpenStudy (anonymous):

You shouldn't need to factor it.\[24x-26=0\]Should be the equation. Move -26 over to get\[24x=26\]Then divide to get the answer.

OpenStudy (anonymous):

:) whoops. Thanks.

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