Question

Find the value of (x^4+y^4 ) and (x^2-y^2)^2, when x^2 + y^2 = 5 and xy = 2.​

Answer 1

GIVEN :-

• x² + y² = 5.
• xy = 2.

TO FIND :-

• value of x⁴ + y⁴.
• value of (x² - y²)².

SOLUTION :-

✮ x⁴ + y⁴,

➳ x² + y² = 5

On squaring both sides we get,

➳ (x² + y²)² = (5)²

➳ (x²)² + (y²)² + 2 × x² × y² = 25

➳ x⁴ + y⁴ + 2x²y² = 25

➳ x⁴ + y⁴ + 2(x × y)² = 25

➳ x⁴ + y⁴ + 2(2)² = 25

➳ x⁴ + y⁴ + 2 × 4 = 25

➳ x⁴ + y⁴ = 25 - 8

➳ x⁴ + y⁴ = 17.

✮ (x² - y²)²,

➵ (x² - y²)²

By using identity :- (a - b)² = a² + b² - 2ab.

➵ (x² - y²)² = (x²)² + (y²)² - 2x²y²

➵ (x² - y²)² = x⁴ + y⁴ - 2x²y²

➵ (x² - y²)² = 17 - 2(xy)²

➵ (x² - y²)² = 17 - 4

➵ (x² - y²)² = 13.

Answer 2

Given

• x² + y² = 5 and xy = 2

We Find

• Value of (x^4 + y^4 ) and (x²- y²) ²

According to the question

▶ x² + y² = 5

On squaring, we get :-

= ( x² + y² )² + = (5)²

= (x²)² + (y²)² + 2 × x² × y² = 25

= x4 + x4 + 2(x × y)² = 25

= x4 + x4 + 2(2)² = 25

= x4 + x4 + 2 × 4 = 25

= x4 + x4 + 8 = 25

= x4 + x4 = 25 - 8

= x4 + x4 = 17

So, therefore x^4 + y^4 is 17

( Identity used :- (a - b)² = a² + b² - 2ab )

= (x² - y²)² = (x²)² + (y²)² - 2xy

= (x² - y²)² = x4 + y4 - 2xy

= (x² - y²)² = 17 - 4

= (x² - y²)² = 13

So, Therefore (x² - y²)² is 13

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