if x+y =8 and xy =15 then find x^2-y^2
Answers
Answered by
5
X+Y=8
SQUARING BOTH SIDES
(X+Y)^2=(8)^2
X^2+Y^2+2XY=64
X^2+Y^2+2×15=64
X^2+Y^2+30=64
X^2+Y^2=64-30
X^2+Y^2=34 ANSWER
SQUARING BOTH SIDES
(X+Y)^2=(8)^2
X^2+Y^2+2XY=64
X^2+Y^2+2×15=64
X^2+Y^2+30=64
X^2+Y^2=64-30
X^2+Y^2=34 ANSWER
Answered by
3
x^2-y^2
=(x+y)*(x-y)
here given only x+y=8
so we have to first find x-y
(x-y)^2=(x+y)^2 - 4xy
=(8)^2 - 4*15
=64 -60
=4
thus, x-y=√4
=2
thus, x^2-y^2
(x+y)*(x-y)
=8*4
=32
=(x+y)*(x-y)
here given only x+y=8
so we have to first find x-y
(x-y)^2=(x+y)^2 - 4xy
=(8)^2 - 4*15
=64 -60
=4
thus, x-y=√4
=2
thus, x^2-y^2
(x+y)*(x-y)
=8*4
=32
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