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Re: Elimination of radials - Confused¿? [#permalink]
Patheinemann wrote:
Actually the question is pretty straight forward, however upon further examination I somehow got confused on the "well-known" fact that anything you do to one side of an algebraic equation, you must do to the other side. I'll try to elaborate with an example:

Given that √(3b-8) = √(12-b), what is b?

I understand that the approach is to cancel both radicals, and then just proceed with a very simple equation. However in strict theory, wouldn't you have to multiply what you do to one side, to the other side as well?

√(3b-8)^2 = √(12-b) * √(3b-8)

I know I am completely wrong, but it is probably only a matter of having too much of this GMAT stuff.
Thanks to anyone that will kindly respond to my inquiry.


For such question to solve, firstly try to form simple equation. For this we need to get out of that square root.
Now, if a=b, then \(a^2 = b^2\)(squaring both sides)
In our case,
\(\sqrt{(3b-8)}^2 = \sqrt{(12-b)}^2\)
-->> \(3b-8=12-b\)
-->> \(4b=24\)
-->> \(b=24/4=6\)
Therefore, b=6
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Elimination of radials - Confused¿? [#permalink]
premnath wrote:
Patheinemann wrote:
Actually the question is pretty straight forward, however upon further examination I somehow got confused on the "well-known" fact that anything you do to one side of an algebraic equation, you must do to the other side. I'll try to elaborate with an example:

Given that √(3b-8) = √(12-b), what is b?

I understand that the approach is to cancel both radicals, and then just proceed with a very simple equation. However in strict theory, wouldn't you have to multiply what you do to one side, to the other side as well?

√(3b-8)^2 = √(12-b) * √(3b-8)

I know I am completely wrong, but it is probably only a matter of having too much of this GMAT stuff.
Thanks to anyone that will kindly respond to my inquiry.


For such question to solve, firstly try to form simple equation. For this we need to get out of that square root.
Now, if a=b, then \(a^2 = b^2\)(squaring both sides)
In our case,
\(\sqrt{(3b-8)}^2 = \sqrt{(12-b)}^2\)
-->> \(3b-8=12-b\)
-->> \(4b=24\)
-->> \(b=24/4=6\)
Therefore, b=6



Doesn't b=5?

\(\sqrt{(3b-8)}^2 = \sqrt{(12-b)}^2\)
-->> \(3b-8=12-b\)
-->> 4b=20
-->> b=20/4
-->> b=5
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Re: Elimination of radials - Confused¿? [#permalink]
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Re: Elimination of radials - Confused¿? [#permalink]
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