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Intern  Joined: 15 Aug 2011
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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.
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Re: Elimination of radials - Confused¿?  [#permalink]

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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.

An equality, you can also raise to some power.
If for two non-zero numbers $$A=B$$, then multiplying both sides by $$A$$ we obtain $$A\cdot{A}=AB$$ but $$A=B$$, so $$A^2=B^2.$$ You can continue and obtain that for any positive integer $$n, \, A^n=B^n.$$

What you have written is true, but it won't help you get rid of the radicals the way it is written.

If two numbers are equal, raised to the same power will remain equal. You do the same to both sides of the equation, you raise them to the second power.
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Re: Elimination of radials - Confused¿?  [#permalink]

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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.

Take some numbers to understand this.
The equality sign means that whatever is on the left hand side, it is equal to whatever there is on the right hand side. So if the LHS is 2, RHS is also 2

2 = 2
Now, you can square both sides and the equality will still hold 2^2 = 2^2
You can raise both sides to any power, the equality will hold.

You can also multiply the same number on both the sides, the inequality will still hold 2*3 = 2*3

You can choose to do whichever operation suits your purpose. Since you want to get rid of the root sign, you would want to square both sides.

Also, notice that here √(3b-8)^2 = √(12-b) * √(3b-8),
√(3b-8) = √(12-b) (even though they don't look same but they are since it is given to us) so √(3b-8)^2 = √(12-b) * √(3b-8) is the same as √(3b-8)^2 = √(12-b)*√(12-b)
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Re: Elimination of radials - Confused¿?  [#permalink]

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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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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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[fraction][/fraction] Elimination of radials - Confused¿?   [#permalink] 15 Jun 2019, 08:02
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