What is a if the 1/m+1/n=1/5? m,n are roots of equation x^2-x-a=0

You need to bring the summation `1/m + 1/n`   to a common denominator, such that:

`1/m + 1/n = (m + n)/(mn)`

Replacing `(m + n)/(mn)` for `1/m + 1/n` in equation `1/m + 1/n = 1/5` , yields:

`(m + n)/(mn) = 1/5`

Since the problem provides the...

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You need to bring the summation `1/m + 1/n`   to a common denominator, such that:

`1/m + 1/n = (m + n)/(mn)`

Replacing `(m + n)/(mn)` for `1/m + 1/n` in equation `1/m + 1/n = 1/5` , yields:

`(m + n)/(mn) = 1/5`

Since the problem provides the information that m and n are the solution to the quadratic equation `x^2 - x - a = 0` , you need to use Vieta's formulas, such that:

`m + n = -(-1)/1 => m + n = 1`

`m*n = -a/1 => m*n = -a`

Replacing 1 for m + n and -a for m*n in equation `(m + n)/(mn) = 1/5` yields:

`1/(-a) = 1/5 => a = -5`

Hence, evaluating the value of coefficient `a` , using Vieta's formulas, under the given conditions, yields `a = -5.`