Question

Given (x+y)=13 and xy=22. Find the value of x2+y2​

Answer 1

Method (I)- x+y=13 and xy=22

It means :- x=11 , y=2

Now for x^2 + y^2 = (11)^2 + (2)^2 = 121+4 = 125

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Method (II)- x^2+y^2 + 2xy = 169
x^2+y^2 = 169 - 2×22
x^2+y^2 = 169 - 44
x^2+y^2 = 125
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Answer 2

$We \: know \: that \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \\ = > {x }^{2} + 2(22) + {y}^{2} = {(13)}^{2} \\ = > \: \: \: \: \: \: {x }^{2} + {y}^{2} + 44 = 169 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + {y}^{2} = 169 - 44 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + {y}^{2} = 125$

Answer 3

$We \: know \: that \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + 2xy + {y}^{2} = {(x + y)}^{2} \\ = > {x }^{2} + 2(22) + {y}^{2} = {(13)}^{2} \\ = > \: \: \: \: \: \: {x }^{2} + {y}^{2} + 44 = 169 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + {y}^{2} = 169 - 44 \\ = > \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {x}^{2} + {y}^{2} = 125$