help solve 1/2X=-1+sqrt(x-2)

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help solve 1/2X=-1+sqrt(x-2)

Mathematics
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correction \[\frac{ 1 }{ 2 }x=-1 +\sqrt{x+2}\]
What you tried yet @Compgroupmail ?
I tried to bring it to the power by \[\frac{ 1 }{ 4 }x=1+(x+2)\] /is this correct?

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see (a+b)^2 is NOT equal to a^2 + b^2 only. \[\large{(\frac{1X}{2})^2 = (-1 + \sqrt{x-2})^2 }\] \[\large{\frac{x^2}{4} = 1 + x-2 - 2\sqrt{x-2} }\]
Oh wait in your corrected question it is x+2 not x-2 so it will be : \[\large{\frac{x} = 1+x-2 -2\sqrt{x+2}}\]
\[\large{\frac{x^2}{4} = 1+x-2 -2\sqrt{x+2}}\]
ok but I still don't understand how you broke it down to that. Can we go through it step by step?
Oh! Sorry I think I confused you, yes why not... so do you know the identity of (a+b)^2 = ?
a^2+2ab+b^2
Right! so in RHS (Right hand side) we have : -1 + sqrt{x+2} , correct?
left side of equation \[(\frac{ 1 }{ 2 }x)^2= \frac{ 1 }{ 4 }x\]. Correct?
right. That is the right hand side. What's next?
Yes , so square the right hand side ... that is : \[\large{( \sqrt{x+2} - 1 )^2 }\] Can you square that?
no. Not 100% how to do it. That's what's messing me up.
Wait for 1 minute, I am opening google chrome as IE is quite slow
ok no problem
add 1 to both sides and square both sides \[(\frac{ x }{ 2 } +1)^2 = x +2\]
\[ \frac{ x^2 }{4 } +x +1 = x+2\]
x^2 =4
ahh. I understand now :) THanks!

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