1.

If a+b+c=0, then roots of the equation `3ax^2+4bx+5c=0` are(A) Positive (B) negative (C) Real and distinct (D) imaginary

Answer» It is given that,
`a+b+c = 0`
`=>b = -(a+c)->(1)`
Now, given equation is,
`3ax^2+4bx+5c = 0`
`:.` Disciminant `(D) = (4b)^2 - 4(3a)(5c) `
`=>D = 16b^2-60ac`
From (1),
`=>D = 16(-(a+c))^2 - 60ac`
`=>D = 16a^2+16c^2+32ac-60ac`
`=>D = 16a^2+16c^2-28ac`
Now, we have two caes.
Case - 1 : When `a` and `c` are of opposite signs.
Then `D` wil be positive.
Case - 2 : When `a` and `c` are of same signs.
`=>D = 16a^2+16c^2-32ac+4ac`
`=>D = 16(a-c)^2+4ac`
Again `D` will be positive.
As `D` is positive, it means roots of the given equation are real and distinct.


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