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If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m
[#permalink]
10 Jul 2024, 01:28

1

Question Stats:

If \(\sqrt{108}= a\sqrt{b}\), where a and b are both positive integers, which of the following could be the value of a+b?

Indicate all such numbers:

A. 9

B. 11

C. 15

D. 20

E. 29

F. 31

Indicate all such numbers:

A. 9

B. 11

C. 15

D. 20

E. 29

F. 31

If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m
[#permalink]
10 Jul 2024, 02:08

Expert Reply

Tricky.

Now, when you practice a question in which you have equality = what you have on the LHS must balance out with the RHS

therefore \(\sqrt{108}\) must be equal to \(a\sqrt{b}\)

Next step is to figure out the factors of \(108=\sqrt{2^2*3^3}\) or \(108=\sqrt{2^2*3^2*3}\)

Therefore a must be a number and \(\sqrt{b}\) must be the numbers or factors of 108 but at the same time we must have the a outside the root and then take the sum

\(a\sqrt{b}\)

\(1*\sqrt{108}\)

\(2*\sqrt{27}\)

\(3*\sqrt{12}\)

\(6*\sqrt{3}\)

\(a\sqrt{b}\)

\(1+108=109\)

\(2+27=29\)

\(3+12=15\)

\(6+3=9\)

I hope this helps

Now, when you practice a question in which you have equality = what you have on the LHS must balance out with the RHS

therefore \(\sqrt{108}\) must be equal to \(a\sqrt{b}\)

Next step is to figure out the factors of \(108=\sqrt{2^2*3^3}\) or \(108=\sqrt{2^2*3^2*3}\)

Therefore a must be a number and \(\sqrt{b}\) must be the numbers or factors of 108 but at the same time we must have the a outside the root and then take the sum

\(a\sqrt{b}\)

\(1*\sqrt{108}\)

\(2*\sqrt{27}\)

\(3*\sqrt{12}\)

\(6*\sqrt{3}\)

\(a\sqrt{b}\)

\(1+108=109\)

\(2+27=29\)

\(3+12=15\)

\(6+3=9\)

I hope this helps

Re: If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m
[#permalink]
10 Jul 2024, 02:31

Carcass wrote:

Tricky.

Now, when you practice a question in which you have equality = what you have on the LHS must balance out with the RHS

therefore \(\sqrt{108}\) must be equal to \(a\sqrt{b}\)

Next step is to figure out the factors of \(108=\sqrt{2^2*3^3}\) or \(108=\sqrt{2^2*3^2*3}\)

Therefore a must be a number and \(\sqrt{b}\) must be the numbers or factors of 108 but at the same time we must have the a outside the root and then take the sum

\(a\sqrt{b}\)

\(1*\sqrt{108}\)

\(2*\sqrt{27}\)

\(3*\sqrt{12}\)

\(6*\sqrt{3}\)

\(a\sqrt{b}\)

\(1+108=109\)

\(2+27=29\)

\(3+12=15\)

\(6+3=9\)

I hope this helps

Now, when you practice a question in which you have equality = what you have on the LHS must balance out with the RHS

therefore \(\sqrt{108}\) must be equal to \(a\sqrt{b}\)

Next step is to figure out the factors of \(108=\sqrt{2^2*3^3}\) or \(108=\sqrt{2^2*3^2*3}\)

Therefore a must be a number and \(\sqrt{b}\) must be the numbers or factors of 108 but at the same time we must have the a outside the root and then take the sum

\(a\sqrt{b}\)

\(1*\sqrt{108}\)

\(2*\sqrt{27}\)

\(3*\sqrt{12}\)

\(6*\sqrt{3}\)

\(a\sqrt{b}\)

\(1+108=109\)

\(2+27=29\)

\(3+12=15\)

\(6+3=9\)

I hope this helps

Hey Carcass,

Could you please explain how did you come up with the highlighted text?

If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m
[#permalink]
10 Jul 2024, 02:47

Expert Reply

the question is a+b

the \(a\) is the value outside of the root and \(b\) is the value INSIDE the root

\(a \sqrt{b}\)

\(a=1\)

\(b=108\)

and so on

the \(a\) is the value outside of the root and \(b\) is the value INSIDE the root

\(a \sqrt{b}\)

\(a=1\)

\(b=108\)

and so on

gmatclubot

If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m [#permalink]

10 Jul 2024, 02:47
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