On factorizing the above equation,(a4 – b4) = [(a2 + b2) (a2 – b2)]The above relation is same as (a + b) (a – b) = a2 – b2Here a = (a2 + b2) and b = (a2 - b2)4x2 – 4a2x + (a4 – b4) = 04x2 – [2(a2 + b2) + 2(a2 – b2)] x + [(a2 + b2) (a2 – b2)] = 04x2 – 2x(a2 + b2) – 2x(a2 – b2) + [(a2 + b2) (a2 – b2)] = 02x[2x - (a2 + b2)] - (a2 - b2) [2x - (a2 + b2)] = 0[2x - (a2 - b2)] [2x - (a2 + b2)] = 0Solving the first part,[2x - (a2 - b2)] = 02x = (a2 - b2)x = (a2 - b2) / 2Solving the second part,[2x - (a2 + b2)] = 02x = (a2 + b2)x = (a2 + b2) / 2

Q. 2 J

Solve the followi

Class 10th Rajasthan Board Mathematics 3. Polynomials

Answer :

On factorizing the above equation,

(a⁴ – b⁴) = [(a² + b²) (a² – b²)]

The above relation is same as (a + b) (a – b) = a² – b²

Here a = (a² + b²) and b = (a² - b²)

4x² – 4a²x + (a⁴ – b⁴) = 0

4x² – [2(a² + b²) + 2(a² – b²)] x + [(a² + b²) (a² – b²)] = 0

4x² – 2x(a² + b²) – 2x(a² – b²) + [(a² + b²) (a² – b²)] = 0

2x[2x - (a² + b²)] - (a² - b²) [2x - (a² + b²)] = 0

[2x - (a² - b²)] [2x - (a² + b²)] = 0

Solving the first part,

[2x - (a² - b²)] = 0

2x = (a² - b²)

x = (a² - b²) / 2

Solving the second part,

[2x - (a² + b²)] = 0

2x = (a² + b²)

x = (a² + b²) / 2

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