Question

x²+y²=13 xy=6,
1) x+y
2)x-y
3) x⁴+y⁴​

Answer 1

Answer:

5,1,97

Step-by-step explanation:

Given

x^2+y^2=13

xy=06

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

(x+y)^2=13+2(6)

(x+y)^2=13+12

(x+y)^2=25

x+y=√25

x+y=5

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

(x-y)^2=13-2(6)

(x-y)^2=13-12

(x-y)^2=1

x-y=√1

x-y=1

(x^2+y^2)^2=x^4+y^4+2x^2y^2

13^2=x^4+y^4+2(xy)^2

169=x^4+y^4+2(6)^2

169=x^4+y^4+2X36

169=x^4+y^4+72

169-72=x^4+y^4

x^4+y^4=97