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If [m][square_root]108[/square_root]= a[square_root]b[/square_root][/m
[#permalink]
10 Jul 2024, 01:28
10 Jul 2024, 01:28
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Question Stats:
71% (01:17) correct
28% (00:53) wrong based on 7 sessions
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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
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
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?
