Question

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

• A. 0
• B. 106
• C. 14
• D. 52

Answer 1

The correct option is A

0

$\text{Given :}{x}^2+y^2=14...\left(1\right)$
$xy=3...\left(2\right)$

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

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

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