Question

if x2+y2=13 and xy=6 find the value of the following. 1.x+y 2.x-y 3.x4+y4

Answer 1

(1)
(x+y)² = x²+y²+2xy

(x+y)² = 13+2(6)

(x+y)² = 13+12

(x+y)² = 25

(x+y) = √25

x+y = 5

(2)
(x-y)² = x²+y²-2xy

(x-y)² = 13-2(6)

(x-y)² = 13-12

(x-y) = √1

x-y = ±1

(3)
x⁴+y⁴ = (x²)²+(y²)²

x⁴+y⁴ = (x²+y²)²-2(x²)(y²)

x⁴+y⁴ = 13²-2(xy)²

x⁴+y⁴ = 169-2(6)²

x⁴+y⁴ = 169-2(36)

x⁴+y⁴ = 169-72

x⁴+y⁴ = 97

Hope it helps....

Answer 2

(x+y)² = x²+y²+2xy

(x+y)² = 13 + 2 x 6

(x+y)² = 25

(x+y) = √25

x+y = 5
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(x-y)² = x²+y² - 2xy

(x-y)² = 13 - 2 x 6 

(x-y)² = 1

(x-y) = √1

x - y = ± 1  [ -1 or +1 ]
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x⁴+y⁴ = (x²+y²)² - 2(x²)* (y²)

x⁴+y⁴ = 13² -  2 * (xy)²

x⁴+y⁴ = 13² - 2 * 36

x⁴+y⁴ = 97

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Hope this helps you !!