SSC CHSL 8 Jan 2017 Shift 3 Paper

Question : 53 of 100

Marks: +1, -0

If x + y = 12 and xy = 32, then what is the value of

x^2+y^2

Solution:

Given :

(x+y)=12

and

xy=32

Using

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

⇒

(12)^2=(x^2+y^2)+(2×32)

⇒

(x^2+y^2)=144-64=80

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