Math, asked by shreyashshobhit0, 14 days ago

1. If x+y = 13, and xy=40 find the value of x-y

Answers

Answered by dkchakrabarty01

0

Answer:

(x-y)^2=x^2+y^2-2xy=x^2+y^2+2xy-4xy=(x+y)^2-4xy

=13^2-4×40

=169-160=9

(x-y)^2=9

x-y=+3 or -3

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