Question

If x^2 + y^2=34 and xy=21/2, find the value of 2(x+y)^2 + (x-y)^2

Answer 1

Given,(x² + y²)=34 ; xy=21÷2
Therefore,
(x+y)²=x²+y²+2xy    [x²+y²=34 and xy=21÷2]
         =34 + 2×21÷2
         =34 + 21
         =55
(x-y)²=x²+y²-2xy
         =34 - 2×21÷2
         =34 - 21
         =13
Now,
  2(x+y)²+(x-y)²
=2×55 + 13
=110+13
=123 (Ans)

Answer 2

If x^2 + y^2=34 and xy=21/2, then the value of 2(x+y)^2 + (x-y)^2 is given as follows:

Given,

(x² + y²)=34 

xy=21/2

We use formulae,

(x + y)² = x² + y² + 2xy

(x - y)² = x² + y² - 2xy

Now, consider,

(x + y)² = x² + y² + 2xy    

from given, we have,

        = 34 + 2 × 21/2

        = 34 + 21

∴ (x + y)² = 55

Now, consider,

(x - y)² = x² + y² - 2xy

from given, we have,

        = 34 - 2 × 21/2

        = 34 - 21

∴ (x - y)² = 13

So, we have,

2(x + y)² + (x - y)²

= 2 × 55 + 13

= 110 + 13

= 123

∴  2(x + y)² + (x - y)² = 123