Question

If x+y=10 and xy =21, find 2(x²+y²)​

Answer 1

We know that.

(x+y)²=x²+y²+2xy

-> x²+y²=(x+y)²-2xy

Given :

x+y=10 and xy=21

Substituting these values...

-> x²+y²=(10)² -2(21)

=100-42=58

Now; Question is to find 2(x²+y²)

=>2(x²+y²)= 2(58)=116

Hope it helps...

Regards;

Leukonov/Olegion

Answer 2

Step-by-step explanation:

$x + y = 10.......(1) \\ xy = 21...........(2) \\from....(1) \\ \: x = 10 - y ......(3) \\ substitute \: eq.(3) \: in \: eq.(2) \\ (10 - y)y = 21 \\ 10y - {y}^{2} - 21 = 0 \\ {y}^{2} - 10y + 21 = 0 \\ {y}^{2} - 7y - 3y + 21 = 0 \\ y(y - 7) - 3(y - 7) = 0 \\ (y - 7) \: and \: (y - 3) \\ y = 7 \: and \: y = 3 \\ substitute \: this \: value \: in \: eq.(3) \\ if \: y = 3 \: then \: \\ x = 10 - 3 = 7 \\ if \: x = 7 \: then \\ x = 10 - 7 = 3 \\ now \: acc. \: to \: question \: \\ 2( {x}^{2} + {y}^{2} ) = {3}^{2} + {7}^{2}\\ = 2(9 + 49) = 2(58)=116$

HOPE IT HELPS YOU❤❤❇❇✌✌