Question

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

Answer 1

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

Answer 2

What you tried yet @Compgroupmail ?

Answer 3

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

Answer 4

see (a+b)^2 is NOT equal to a^2 + b^2 only.

\[\large{(\frac{1X}{2})^2 = (-1 + \sqrt{x-2})^2 }\]
\[\large{\frac{x^2}{4} = 1 + x-2 - 2\sqrt{x-2} }\]

Answer 5

Oh wait in your corrected question it is x+2 not x-2

so it will be : \[\large{\frac{x} = 1+x-2 -2\sqrt{x+2}}\]

Answer 6

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

Answer 7

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

Answer 8

Oh! Sorry I think I confused you, yes why not...

so do you know the identity of (a+b)^2 = ?

Answer 9

a^2+2ab+b^2

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

ok no problem

Answer 17

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

Answer 18

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

Answer 19

x^2 =4

Answer 20

ahh. I understand now :) THanks!