Math, asked by sonaldalal31, 1 year ago

x+y=_4andxy=2,then find the value of x^2+y^2​

Answers

Answered by sujayraj24
1

Answer:

(x+y)²=x²+y²+2xy

(-4)²=x²+y²+2*2

16=x²+y²+4

12=x²+y²

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theprimus2446: bro it was x^2 +y^2
theprimus2446: not (x+y)^2
Answered by theprimus2446
0

we know that,

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

so, now...

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

=(x+y)(x^2 - xy +y^2)=16

= 4(x^2 -2+ y^2)=16

= x^2 + y^2 -2 =16-4

=x^2 + y^2 =12+2

=x^2 + y^2 = 14

L.H.S. = R.H.S.

HECNCE VERIFIED


theprimus2446: but I made a mistake
theprimus2446: I will correct it in the comment section
theprimus2446: 4(x^2+y^2 -2) = 16
theprimus2446: x^2+y^2-2=16÷4
theprimus2446: x^2 + y^2 = 4+2
theprimus2446: x^2 + y^2 = 6
theprimus2446: this is correct
theprimus2446: (I think)
theprimus2446: this is the correct one
theprimus2446: I checked it
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