anonymous
  • anonymous
Solving Square Root Equations 5(squareroot of)x-1=(squaredrootof) x+1
Mathematics
schrodinger
  • schrodinger
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dumbcow
  • dumbcow
square both sides --> 25(x-1) = x+1
anonymous
  • anonymous
yes, but i need to get x. :(
dumbcow
  • dumbcow
you can solve it from there get x by itself

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anonymous
  • anonymous
it would be 25x? thats somethign i dont get.
anonymous
  • 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\[x=\]
anonymous
  • anonymous
i dont get the 5th step...
anonymous
  • anonymous
The "-x +25 -x +25" step?
anonymous
  • anonymous
wouldnt it be 24x-24=0? yeah yhat step.
anonymous
  • anonymous
No\[-25-1=-26\]
anonymous
  • anonymous
wooow... feeling like a genius...
anonymous
  • 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.
anonymous
  • anonymous
I hate making those simple mistakes.
anonymous
  • anonymous
okay. so, if i also take the x out, because i want to have =0, it owuld be 24x-26=o, correct?
anonymous
  • anonymous
Yes.
anonymous
  • anonymous
so now i factor it, right?
anonymous
  • 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.
anonymous
  • anonymous
:) whoops. Thanks.

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