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Question
`3a^2x^2+8abx+4b^2=0`
Solution
Given:
`3a^2x^2+8abx+4b^2=0`
On comparing it with` Ax^2+Bx+C=0` we get:
`Aa=3a^2, B=8ab and C=4b^2`
Discriminant D is given by:
`D=(B^2-4AC) `
=`(8ab)^2-4xx3a^2xx4b^2`
=` 16a^2b^2>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α =(-b+sqrt(D))/(2a)=(-8ab+sqrt16a^2b^2)/(2xx3a^2)=(-8ab+4ab)/(6a^2)=(-4ab)/(6a^2)=(-2b)/(3a) `
`β=(-b-sqrt(D))/(2a)=(-8ab- sqrt16a^2b^2)/(2xx3a^2)=(-8ab-4ab)/(6a^2)=(-12ab)/(6a^2)=(-2b)/(a)`
Thus, the roots of the equation are `(-2b)/(3a) and (-2b)/a`
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