Question

If 2(x+y)^2=116, and xy= 24, find the value of x^2+y^2.

Answer 1

Answer:

10

Step-by-step explanation:

$2(x + y)^{2} = 116 \\ {(x + y)}^{2} = \frac{116}{2} \\ {(x + y)}^{2} = 58....(1) \\ xy = 24.....(2) \\ \because \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \therefore \: 58 = {x}^{2} + {y}^{2} + 2 \times 24 \\ \therefore \: 58 = {x}^{2} + {y}^{2} + 48 \\ \therefore \: {x}^{2} + {y}^{2} = 58 - 48 \\ \huge \purple{ \boxed{\therefore \: {x}^{2} + {y}^{2} = 10}}$

Answer 2

Answer:

10

Step-by-step explanation:

2(x+y)^2=116

xy= 24

x^2+y^2 =?

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x^2+y^2= (x+y)^2- 2xy

• (x+y)^2 = 116/2 = 58
• x^2+y^2 = 58- 2*24 = 58 - 40= 10