Question

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

Answer 1

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

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Answer 2

Question:

Find the value of $2(x+y)^2 + (x-y)^2$.

To find the value, we will first expand both brackets.

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$2(x+y)^2 + (x-y)^2=[2(x^2+y^2)+4xy]+[(x^2+y^2)+2xy]$

$2(x+y)^2 + (x-y)^2=3(x^2+y^2)+6xy$

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Substituting the value above, $x^2 + y^2=34$ and $xy=\dfrac{21}{2}$

we get $3\times34+6\times\dfrac{21}{2}$

which is $102+63=165$.