Question

11. If x^2+ y^2 = 13 and x*y = 6, find the value of the following.
i. x+y
ii. X-y
iii. x^4 + y^4​

Answer 1

Answer:

ets assume,

x

1

=a,

y

1

=b, then the above equations reduces to

2a+6b=13 ...(1)

3a+4b=12 ...(2)

Multiplying (1) by 3 and (2) by 2, we get

6a+18b=39 ...(3)

6a+8b=24 ...(4)

Subtracting (3) and (4), we get

10b=15⇒b=

2

3

Now, 2a=13−6×

2

3

=13−9=4

⇒a=2