If $a,b,c \in R$ and $a+b+c=0$, then the quadratic equation $4ax^{2}+3bx+2c=0$ has
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one positive and one negative
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imaginary roots
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real roots
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none of these
Discriminant D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac. Since a+b+c=0, b = -(a+c). D = 9(a+c)^2 - 32ac = 9a^2 + 18ac + 9c^2 - 32ac = 9a^2 - 14ac + 9c^2. This is always positive for real a, c (as it's a quadratic in a/c with negative discriminant), so roots are real.
To determine the nature of the roots, we evaluate the discriminant of the quadratic equation, which is defined as b squared minus 4ac. For the equation 4ax squared plus 3bx plus 2c equals 0, the discriminant is (3b) squared minus 4 times 4a times 2c, simplifying to 9b squared minus 32ac. Substituting c equals negative a minus b into the expression gives 9b squared minus 32a times negative a minus b, which expands to 9b squared plus 32a squared plus 32ab. We can write this as 16a squared plus 16ab plus 4b squared plus 16a squared plus 16ab plus 5b squared, which factors into (4a plus 2b) squared plus 16a times (a plus b) plus 5b squared. Substituting a plus b equals negative c gives (4a plus 2b) squared minus 16ac plus 5b squared, which can be rearranged to (4a plus 2b) squared plus 4ac plus 4b squared plus b squared minus 20ac. This sum of squares is always non-negative, guaranteeing that the quadratic equation has real roots.