Answer :

**Given:** equation 4x^{2} – 4a^{2}x + (a^{4}-b^{4}) =0

**To find:** The roots by factorisation method

**Method Used:**

To solve the quadratic equation by factorisation method, follow the steps:

1) Multiply the coefficient of x^{2} and constant term.

2) factorise the result obtained in step 1.

3) Now choose the pair of factors in such a way that after adding or subtracting(splitting) them

You get coefficient of x.

**Explanation:**

Here,

Coefficient of x^{2} = 4

Constant term = (a^{4}-b^{4})

As a^{2}-b^{2} = (a-b) (a +b)

(a^{4}-b^{4}) = (a^{2} – b^{2}) (a^{2}+b^{2})

As 4(a^{4}-b^{4}) = 4[(a^{2} – b^{2}) (a^{2}+b^{2})]

Coefficient of x = - 4a^{2}

=- 2a^{2} - 2a^{2}

Add and subtract 2b^{2} to get,

= -2a^{2} - 2a^{2} + 2b^{2} - 2b^{2}

= -2a^{2} – 2b^{2} – 2a^{2} + 2b^{2}

= - [2(a^{2}+ b^{2}) + 2(a^{2}- b^{2})]

∴ 4x^{2} – 4a^{2}x + (a^{4}-b^{4}) = 0

⇒ 4x^{2} - [2(a^{2}+ b^{2}) + 2(a^{2}- b^{2})] x + (a^{2} – b^{2}) (a^{2}+b^{2}) = 0

⇒ 4x^{2} - 2(a^{2}+ b^{2}) x - 2(a^{2}- b^{2}) x + (a^{2} – b^{2}) (a^{2}+b^{2}) = 0

⇒ [4x^{2} - 2(a^{2}+ b^{2}) x] – [2(a^{2}- b^{2}) x - (a^{2} – b^{2}) (a^{2}+b^{2})] = 0

⇒ 2x [2x-(a^{2}+ b^{2})] - (a^{2} – b^{2}) [2x-(a^{2}+ b^{2})] = 0

⇒ [2x-(a^{2}+ b^{2})] [2x-(a^{2}- b^{2})] = 0

⇒ 2x-(a^{2}+ b^{2}) = 0 and 2x-(a^{2}- b^{2}) = 0

Hence roots are .

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