Question

35.) If x+y=12 and xy=32 find the
value of x²+y²?​

Answer 1

Answer:

x+y = 12

(x+y)² = 12²

x²+y²+2xy = 144

x²+y²+2*32 = 144

x²+y² = 144 - 64

x²+y² = 80

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Answer 2

Answer:

(x+y)²=x²+y²+2xy

12²=x²+y²+2×32

144=x²+y²+64

144-64=x²+y²

x²+y²=80