Question

If x+y=4. And xy=6. Find x²+y² *

Answer 1

Answer:

$x^{2} +y^{2} = 4$

Step-by-step explanation:

x + y = 4    

[on squaring both sides }

$(x+y)^{2} = 4^{2}$

$x^{2} +y^{2} + 2xy = 16$

$x^{2} + y^{2} + 2(xy) = 16$               [ xy = 6]

$x^{2} + y^{2} + 2(6) = 16$

$x^{2} +y^{2} + 12 = 16$

$x^{2} +y^{2} = 16 - 12\\x^{2} + y^{2} = 4$

PLEASE MARK IT AS BRAINLIEST

Answer 2

Answer:

4 .

Step-by-step explanation:

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$x + y = 4 \\ xy = 6 \\ = > {(x + y)}^{2} = {(4)}^{2} \\ = > {x}^{2} + {y}^{2} + 2xy = 16 \\ = > {x}^{2} + {y}^{2} +2 (6) = 16 \\ = > {x}^{2} + {y}^{2} = 16 - 12 \\ = > {x}^{2} \: \: \: + \: \: \: {y}^{2} \: \: = \: \: 4$