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Re: If srq(12+6*3^1/2)= a^1/2+b^1/2 and a-b=6 [#permalink]
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mikej wrote:
Please share OA


\(12+6\sqrt{4}=12+2\sqrt{27}\)
We have
27=9*3
27=27*1
Since: 9-3=6 => a=9, b=3, a+b=12
In detail:
\(12+2\sqrt{27} \)
\(=9+3+2\sqrt{9*3}=(\sqrt{9}+\sqrt{3})^2\)
Equivalent to:
\(a+b+2\sqrt{a*b}=(\sqrt{a}+\sqrt{b})^2\)
=>a+b=9+3=12

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Re: If srq(12+6*3^1/2)= a^1/2+b^1/2 and a-b=6 [#permalink]
I spent more than ten minutes on this. Is that a lot, or is it normal? The correct answer is: a + b = 12.
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Re: If srq(12+6*3^1/2)= a^1/2+b^1/2 and a-b=6 [#permalink]
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107John117 wrote:
I spent more than ten minutes on this. Is that a lot, or is it normal? The correct answer is: a + b = 12.




:shock: :shock:
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Re: If srq(12+6*3^1/2)= a^1/2+b^1/2 and a-b=6 [#permalink]
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Carcass wrote:
107John117 wrote:
I spent more than ten minutes on this. Is that a lot, or is it normal? The correct answer is: a + b = 12.




:shock: :shock:

:D
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Re: If srq(12+6*3^1/2)= a^1/2+b^1/2 and a-b=6 [#permalink]
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