Question

Q.b ifx4+x2y2+y4= 21 and x² + xy + y2 = 7, then the Value of (1/x2+1/y2)is
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Answer 1

Answer:

| If x4 + x2y2 + y4 = 21, and x2 + xy + y2 = 3, then what is the value of 4xy?

A. -8

B. 4

C. -4

D. 12

Answer 2

Step-by-step explanation:

Given:

If x4 + x2y2 + y4 = 21 and x2 + xy + y2 = 7

Formula used:

x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2)

Calculation:

x4 + x2y2 + y4 = 21 and

x2 + xy + y2 = 7 --- (1)

x4 + x2y2 + y4 = (x2 – xy + y2) (x2 + xy + y2)

⇒ 21 = (x2 - xy + y2) × 7

⇒ (x2 - xy + y2) = 21/7

⇒ (x2 - xy + y2) = 3 --- (2)

equation (1) from equation (2)

2xy = 4

⇒ xy = 2

⇒ (xy)2 = 4

From equation (1)

x2 + y2 = 7 – 2

⇒ x2 + y2 = 5

Now,

(1/x2 + 1/y2)

⇒ (x2 + y2)/(xy)2

⇒ 5/4

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