If sum of two numbers is #7# and their product is #-64#, find the numbers?

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Answer 1 · 2017-10-03T16:25:31

Numbers are #(7-sqrt305)/2# and #(7+sqrt305)/2#

As #x+y=7#, #y=7-x# and

#x(7-x)=-64# i.e. #7x-x^2=-64#

or #x^2-7x-64=0#

Using quadratic formula #x=(-(-7)+-sqrt((-7)^2-4*1*(-64)))/2#

= #(7+-sqrt(49+256))/2#

= #(7+-sqrt305)/2#

Observe that if #x=(7-sqrt305)/2#, #y=(7+sqrt305)/2#

and if #x=(7+sqrt305)/2#, #y=(7-sqrt305)/2#

Hence numbers are #(7-sqrt305)/2# and #(7+sqrt305)/2#