Question

if x+y =10 and xy=21 find 2(x^2+y^2)​

Answer 1

Answer:

X² + y² + 2xy = (x +y)²

=> x² + y² + 2×21 = (10)²

=> x² + y² + 42 = 100

=> x² + y² = 100 - 42

=> x² + y² = 58 ans

Answer 2

Step-by-step explanation:

X + Y = 10

XY = 21

Y = 21/x

X + 21/x = 10

X^2 + 21 = 10x

X^2 - 10x + 21 = 0

X^2 - 7x - 3x + 21 = 0

X ( x - 7 ) - 3 ( x - 7 ) = 0

(x - 7) ( x - 3) = 0

X = 7, 3

XY = 21

Y = 21/x

1st : x = 3, y = 21/3 => y = 7

2nd : x = 7, y = 21/7 => y = 3

2( x^2 + y^2)

Let x = 3 and y = 3

2( 3^2 + 3^2)

2 ( 9 + 9) => 2(18) => 36

Let x = 7 and y = 7

2 ( 7^2 + 7^2)

2( 49 + 49) => 2 ( 98) => 196

If u want u can Change the value in 2(x^2 + y^2)

Hope it helps you

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