roots (2)

If I were to see problem like that, I'd immediately grab calculator. Dealing with sums and square roots is a b!tch.

Does anyone know how to solve this one w/o a calculator?

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by Ian Stewart » Mon May 04, 2009 12:55 pm

We can use the combined rates formula:

1/T = 1/A + 1/B

where T is the time together, A is the time for A alone, and B is the time for B alone. According to the information in the question, A alone takes sqrt(8) + sqrt(7) hours, while B alone takes sqrt(7) + sqrt(6) hours. So we need to solve the following for T:

1/T = 1/[sqrt(8) + sqrt(7)] + 1/[sqrt(7) + sqrt(6)]

We have sums with square roots in the denominators of fractions, so we certainly want to 'rationalize' the denominators here - that is, get rid of the roots. We use the 'difference of squares' pattern:

1/T = (1/[sqrt(8) + sqrt(7)])*([sqrt(8) - sqrt(7)]/[sqrt(8) - sqrt(7)]) + (1/[sqrt(7) + sqrt(6)])*([sqrt(7) - sqrt(6)] / [sqrt(7) - sqrt(6)] )

1/T = ([sqrt(8) - sqrt(7)])/(8 - 7) + ([sqrt(7) - sqrt(6)])/(7-6)

1/T = sqrt(8) - sqrt(7) + sqrt(7) - sqrt(6)
1/T = sqrt(8) - sqrt(6)
T = 1/[sqrt(8) - sqrt(6)]

Again we need to use the 'difference of squares' to get the roots out of the denominator:

T = (1/[sqrt(8) - sqrt(6)])*([sqrt(8) + sqrt(6)] /[sqrt(8) + sqrt(6)] )
T = (sqrt(8) + sqrt(6))/2

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