Question

2 If x + y = 10 and xy = 21, find 2(x2 + y2).​

Answer 1

Answer :

x+y=10...(1) and xy=21=>x=21/y...(2) Substitute (1) in (2) then we get

21/y +y=10=>

${y}^{2} + 21 = 10y = > {y}^{2} - 10y + 21 =$

$0 = > {y}^{2} - 7y - 3y + 21 = 0 = >$

$y(y - 7) - 3(y - 7) = 0 = >$

$(y - 3)(y - 7) = 0$

So y=3 or 7 When y=3, x+3=10 => x=7

When y=7, x+7=10=>x=3 So

$2( {x}^{2} + {y}^{2}) = 2( {3}^{2} + {7}^{2} ) = 2(9 + 49)$

So the value of 2(x2+y2) is 116. Hope it helps you.