$If x^{2} + y^{2} = 14 and x y = 3,$
then find the value of
$2 (x + y)^{2} - 5 (x - y)^{2} .$

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Solution

The correct option is A
0

$Given : x^{2} + y^{2} = 14 . . . (1)$
$x y = 3 . . . (2)$

Using the identities,
$(a + b)^{2} = a^{2} + 2 a b + b^{2}$
$and (a - b)^{2} = a^{2} - 2 a b + b^{2}$ ,

$2 (x + y)^{2} - 5 (x - y)^{2}$

$= 2 (x^{2} + y^{2} + 2 x y) - 5 (x^{2} - 2 x y + y^{2})$
$= 2 x^{2}$ + $2 y^{2}$ + $4 x y$ - $5 x^{2}$ + $10 x y - 5 y^{2}$
$= (2 x^{2} - 5 x^{2}) + (2 y^{2} - 5 y^{2}) + (4 x y + 10 x y)$
$= - 3 x^{2} - 3 y^{2} + 14 x y$
$= - 3 (x^{2} + y^{2}) + 14 x y$
(Substituting values from (1) and (2))
$= (- 3 \times 14) + (14 \times 3)$
$= - 42 + 42$
$= 0$

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Similar questions

If $x^{2}$ + $y^{2}$ = 14 and xy = 3, find the value of $2 {(x + y)}^{2} - 5 {(x - y)}^{2}$ .

If $x^{2} + y^{2} = 14$ and $x y = 3$ , find the value of $2 (x + y)^{2} - 5 (x - y)^{2}$ .

$If x^{2} + y^{2} = 14 and x y = 3,$
then find the value of
$2 (x + y)^{2} - 5 (x - y)^{2} .$

x^{2} + y^{2} - x y = 3

y - x = 1

\frac{x y}{x^{2} + y^{2}}

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