Given (x+y) = 14 and xy = 54, find the value of x^2–y^2 .
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(x²-y²)= (x+y) ( x-y)
we know the value of x+ y
so xy= 54
y= 54/ x
x+ (54)/x= 14
x²+54. = 14x
x²-14x + 54= 0
x= 8,6
when ....x= 6 ....y= 9
x²-y² = 14 × -3
= -42
we know the value of x+ y
so xy= 54
y= 54/ x
x+ (54)/x= 14
x²+54. = 14x
x²-14x + 54= 0
x= 8,6
when ....x= 6 ....y= 9
x²-y² = 14 × -3
= -42
meethi1234:
how x²-y² =14×-3????
i think it should be 36-81=-45
as x + y = 14 abd x-y= -3 ....so their multiplication is x²-y²
oh yess
i m.srry
no its fine
Answered by
0
x^2 -x^2 =(x+y)(x-y)
lets find (x-y)
(x+y)^2-(x-y)^2=4xy
=14^2-(x-y)^2=4×54
=196-(x-y)^2=216
=196-216=(x-y)^2
=√-20=(x-y)
now
(x+y)(x-y)
=14 ×√-20
=14√-20
lets find (x-y)
(x+y)^2-(x-y)^2=4xy
=14^2-(x-y)^2=4×54
=196-(x-y)^2=216
=196-216=(x-y)^2
=√-20=(x-y)
now
(x+y)(x-y)
=14 ×√-20
=14√-20
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