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Question
`4x^2-4a^2x+(a^4-b^4)=0`
Solution
We write, `-4a^2x=-2(a^2+b^2)x-2(a^2-b^2)x as`
`4x^2xx(a^4-b^4)=4(a^4-b^4)x^2=[-2(a^+b^2)]x xx[-2(a^2-b^2)]x`
∴` 4x^2-4a^2x(a^4-b^4)=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^+b^2)][2x-(a^2-b^2)]=0`
⇒` 2x-(a^2+b^2)=0 or 2x-(a^2-b^2)=0`
⇒`x=(a^2+b^2)/2` or `(a^2-b^2)/2`
Hence, `(a^2+b^2)/2` and `(a^2-b^2)/2`are the roots of the given equation.
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