Question

If x²+y²=34 and xy=10 whole 1/2 find the value of 2(x+y)²+(x-y)²​

Answer 1

Solution:

It is given that,

x²+y² = 34 ----(1)

xy = 10½ = 21/2 ---(2)

Value of 2(x+y)²+(x-y)²

= (x+y)² + [(x+y)²+(x-y)²]

= x²+y²+2xy + 2(x²+y²)

= 34 + 2×(21/2)+2×34

= 34 + 21 + 68

= 123

•••

Answer 2

$x^{2} +y^{2} = 34\\xy = 10\frac{1}2}=\frac{21}{2} \\we know,\\(x+y)^{2} = x^{2} +2xy+y^{2}$

             = $x^{2} +y^{2} +2$×$\frac{21}{2}$

             = 34 + 21

             =55

$(x-y)^{2}=x^{2} -2xy+y^{2}$

             = $x^{2} +y^{2}-2$×$\frac{21}{2}$

             =34-21

             =13

Now,

$2(x+y)^{2}+(x-y)^{2}$

= 2×55 + 13

= 110+13

=123