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14 Apr 2010, 23:05
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This question is from OG 11 pg,22 Q16,
Can someone please help me how you solve the problem, the way it is explained(on OG answer)..
If \sqrt{3-2x}=\sqrt{2x}+1, then 4xΛ2=
A:6x-1
In the answer it says square both side of given equation so,
\sqrt{3-2x}=\sqrt{2x}+1
will become
(\sqrt{3-2x})Λ2=(\sqrt{2x}+1)Λ2
I dont understand why you square at this point.(I know how to work from this point)
Do you square both side because we are trying to find 4xΛ2 which is square..
Say, if it was only 4x, then do we not square to work the problem?
Can someone explain me the principle to this?!
Thanks in advance...
Can someone please help me how you solve the problem, the way it is explained(on OG answer)..
If \sqrt{3-2x}=\sqrt{2x}+1, then 4xΛ2=
A:6x-1
In the answer it says square both side of given equation so,
\sqrt{3-2x}=\sqrt{2x}+1
will become
(\sqrt{3-2x})Λ2=(\sqrt{2x}+1)Λ2
I dont understand why you square at this point.(I know how to work from this point)
Do you square both side because we are trying to find 4xΛ2 which is square..
Say, if it was only 4x, then do we not square to work the problem?
Can someone explain me the principle to this?!
Thanks in advance...
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14 Apr 2010, 23:24
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hi,
this can also be done as under
sqrt{3-2x}=sqrt{2x}+1
=> sqrt{3-2x} - sqrt{2x} = 1
squaring both sides now, we get
=> {3-2x} + {2x} - 2*sqrt{3-2x}{2x} = 1
=> 3 - 2*sqrt{3-2x}{2x} = 1
=> sqrt{3-2x}{2x} = 1
squaring both sides again
{3-2x}{2x} = 1
6x - 4x^2 = 1
which gives 4x^2 = 6x-1
hope i am clear.
this can also be done as under
sqrt{3-2x}=sqrt{2x}+1
=> sqrt{3-2x} - sqrt{2x} = 1
squaring both sides now, we get
=> {3-2x} + {2x} - 2*sqrt{3-2x}{2x} = 1
=> 3 - 2*sqrt{3-2x}{2x} = 1
=> sqrt{3-2x}{2x} = 1
squaring both sides again
{3-2x}{2x} = 1
6x - 4x^2 = 1
which gives 4x^2 = 6x-1
hope i am clear.
14 Apr 2010, 23:26
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gmatJP
This question is from OG 11 pg,22 Q16,
Can someone please help me how you solve the problem, the way it is explained(on OG answer)..
_____________
I dont understand why you square at this point.(I know how to work from this point)
Do you square both side because we are trying to find 4xΛ2 which is square..
Say, if it was only 4x, then do we not square to work the problem?
Can someone explain me the principle to this?!
Thanks in advance...
we square both the sides simply to get rid of the square roots (surd).Can someone please help me how you solve the problem, the way it is explained(on OG answer)..
_____________
I dont understand why you square at this point.(I know how to work from this point)
Do you square both side because we are trying to find 4xΛ2 which is square..
Say, if it was only 4x, then do we not square to work the problem?
Can someone explain me the principle to this?!
Thanks in advance...
there is no other principle to this
