Question

$x + y = 10 \: and \: xy = 21 \: find \: the \: value \: of \: 2(x^{2} + y^{2} )$

Answer 1

$(x + y) = 10 - - - (1)\\ xy = 21 \\ have \: to \: find \: 2( {x}^{2} + {y}^{2} ) \\ \\ in \: eq.1 \\ (x + y) = 10 \\ squaring \: both \: sides \\ {(x + y)}^{2} = 100 \\ {x}^{2} + {y}^{2} + 2xy = 100 \\ {x}^{2} + {y}^{2} = 100 - 2xy \\ (xy = 21) \\ {x}^{2} + {y}^{2} = 100 - 42 \\ {x}^{2} + {y}^{2} = 58 \\ \\ so \\ 2( {x}^{2} + {y}^{2} ) = 2 \times 58 \\ so \\ \\2( {x}^{2} + {y}^{2} ) = 116$

Answer 2

$\bf\huge\color{Red}{Hello\:User}$

(x+y) = 10
xy = 21
so,
2(x2 + y2)
= 2( x + y )^2 - 4xy
= 2( 10 )^2 - 4(21)
= 200 - 84
= 116