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Question
`4x^2-4a^2x+(a^4-b^4)=0`
Solution
The given equation is `4x^2-4a^2x+(a^4-b^4)=0`
Comparing it with `Ax^2+Bx+C=0`
`A=4,B=-4a^2 and C=a^4-b^4`
∴ Discriminant,` B^2-4AC=(-4a^2)^2-4xx4xx(a^2-b^4)=16a^4-16a^4=16b^4=16b^4>0`
So, the given equation has real roots
Now, `sqrtD=sqrt16b^4=4b^2`
∴`α = (-B+sqrt(D))/(2A)=(-(-4a^2)+4b^2)/(2xx4)=(4(a^2+b^2))/8=(a^2+b^2)/2`
`β= (-B-sqrt(D))/(2A)=(-(-4a^2)-4b^2)/(2xx4)=(4(a^2-b^2))/8=(a^2-b^2)/2`
Hence, `1/2(a^2+b^2)` and `1/2(a^2-b^2)` are the roots of the given equation.
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