Compgroupmail
help solve 1/2X=1+sqrt(x2)



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Compgroupmail
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correction \[\frac{ 1 }{ 2 }x=1 +\sqrt{x+2}\]

SheldonEinstein
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What you tried yet @Compgroupmail ?

Compgroupmail
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I tried to bring it to the power by \[\frac{ 1 }{ 4 }x=1+(x+2)\]
/is this correct?

SheldonEinstein
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see (a+b)^2 is NOT equal to a^2 + b^2 only.
\[\large{(\frac{1X}{2})^2 = (1 + \sqrt{x2})^2 }\]
\[\large{\frac{x^2}{4} = 1 + x2  2\sqrt{x2} }\]

SheldonEinstein
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Oh wait in your corrected question it is x+2 not x2
so it will be : \[\large{\frac{x} = 1+x2 2\sqrt{x+2}}\]

SheldonEinstein
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\[\large{\frac{x^2}{4} = 1+x2 2\sqrt{x+2}}\]

Compgroupmail
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ok but I still don't understand how you broke it down to that. Can we go through it step by step?

SheldonEinstein
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Oh! Sorry I think I confused you, yes why not...
so do you know the identity of (a+b)^2 = ?

Compgroupmail
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a^2+2ab+b^2

SheldonEinstein
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Right! so in RHS (Right hand side) we have : 1 + sqrt{x+2} , correct?

Compgroupmail
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left side of equation \[(\frac{ 1 }{ 2 }x)^2= \frac{ 1 }{ 4 }x\].
Correct?

Compgroupmail
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right. That is the right hand side. What's next?

SheldonEinstein
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Yes , so square the right hand side ... that is :
\[\large{( \sqrt{x+2}  1 )^2 }\]
Can you square that?

Compgroupmail
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no. Not 100% how to do it. That's what's messing me up.

SheldonEinstein
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Wait for 1 minute, I am opening google chrome as IE is quite slow

Compgroupmail
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ok no problem

Algebraic!
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add 1 to both sides and square both sides
\[(\frac{ x }{ 2 } +1)^2 = x +2\]

Algebraic!
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\[ \frac{ x^2 }{4 } +x +1 = x+2\]

Algebraic!
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x^2 =4

Compgroupmail
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ahh. I understand now :) THanks!