Question

If x+y=2 & xy=1, then find x^4+y^4.​

Answer 1

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Answer 2

Answer:

The required answer is 2

step-by-step explanation:

${x}^{4}$ + ${y}^{4}$

= ( ${x}^{4}$ + ${y}^{4}$ + 2${x}^{2}$${y}^{2}$ ) - 2${x}^{2}$${y}^{2}$

= ${(x^2+y^2)}^{2}$-2${x}^{2}$${y}^{2}$

= ${((x+y)^2-2xy)}^{2}$ -2 ${(xy)}^{2}$

Now, it is given that

x+y = 2

and

xy = 1

Putting these values in ${eq}^{n}$ 1

we get,

= ${(2^2-2×1)}^{2}$ - 2 × ${1}^{2}$

= ${(4-2)}^{2}$ - 2

= ${2}^{2}$ - 2

= 4 - 2

= 2