Find x.

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|dw:1342256143778:dw|
Try by rationalization process...
|dw:1342256275514:dw|

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Other answers:

Multiply and divide both the sides by: \[\sqrt{x+4}-\sqrt{x -3}\]
You mean multiply by \[(\sqrt{x+4} - \sqrt{x-3})/(\sqrt{x+4} - \sqrt{x-3})\]
Yes...
check @georgeblue22 once.. its very simple
|dw:1342256415787:dw|
i mean check @georgeblue22 reply once
No no don't do this...
I don't understand @georgeblue22's reply.
just do what @georgeblue22 said.. I was thinking it as an expression but it is an equation.. I forgot to see there 1 on right hand side...
\[\Huge{\color{gold}{\star \star}{\color{orange}{\text{Rationlise}}}}\]
best hint
Oh yes, he brought the denominator to the other side.
See, multiply both the sides by : \(\sqrt{x+4}+\sqrt{x -3}\)..
That will help to get of the square root on the left side.
Understood @pratu043 ??? :)
You will left with only one square root term.... after solving..
Left side will be: |dw:1342256668499:dw|
No you are doing wrong..
What's the mistake?
|dw:1342256752422:dw|
Did you try squaring on both the sides ?
See show me what you get after multiply the term I said above... Do it slowly...
Which term?
So 7 = sqrt(x + 4) + sqrt(x - 3).
\[\frac{(\sqrt{x+4} - \sqrt{x-3})}{(\sqrt{x+4} + \sqrt{x-3})} \times (\sqrt{x+4} + \sqrt{x-3}) = 1 \times (\sqrt{x+4} + \sqrt{x-3})\] Now solve this..
You will be left with: \[\sqrt{x+4}- \sqrt{x-3} = \sqrt{x+4} + \sqrt{x-3}\] Now look it properly what is the term that is getting cancelling on both the sides???
sqrt(x + 4).
Yes then cancel it and tell me what you are left with now..
|dw:1342257083455:dw|
Can you bring them on one side now??
|dw:1342257134878:dw|
Yes add them...
|dw:1342257180914:dw|
Yes... Have faith you are in a right direction.. Now squaring both the sides what will you get??
sqaure both sides @pratu043
|dw:1342257244043:dw|
And then put the brackets = 0 and find x from here..
|dw:1342257276476:dw|
Yes.. You are right...
yes @pratu043 u got it ;) deserve a medal ;)
Thanks!!!
yw ;)
See @pratu043 if we are given with: \[\large (x-1)(y-2) = 0\] then you can do directly this: \[(x - 1) =0 \quad and \quad (y - 2) = 0\] Getting???
So at the end we are left with: \[4(x-3) = 0\] As \(4 \ne 0\) So, You can put directly: \[(x-3) = 0\] \[x = 3\]
Oh I see .... I never knew that.
But this is possible in case we have 0 in right hand side.. Be careful...

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