If $a + b + c = 0$ then the roots of the equation $4 a x ^ { 2 } + 3 b x+ 2 c = 0$ where $a , b , c \in R$ are
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real and distinct
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imaginary
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real and equal
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infinite
The discriminant of 4ax^2 + 3bx + 2c = 0 is D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac. Given a+b+c=0, b=-(a+c). Substitute: 9(a+c)^2 - 32ac = 9(a^2 + 2ac + c^2) - 32ac = 9a^2 - 14ac + 9c^2. This quadratic in a/c has a negative discriminant (14^2 - 4*9*9 = 196 - 324 < 0), meaning it is always positive for real a, c. Thus D > 0, so roots are real and distinct.
To determine the nature of the roots, we evaluate the discriminant D = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac. Because a, b, and c are in real numbers and a + b + c = 0, we know that b = -(a+c). Substituting this into the discriminant gives D = 9(a+c)^2 - 32ac = 9a^2 - 14ac + 9c^2. We can rewrite this as 9(a - c)^2 + 4ac, which cannot be properly evaluated for all real numbers to guarantee real and distinct roots, but testing with a=1, b=-2, c=1 yields D = 36 - 32 = 4, which is greater than zero, resulting in real and distinct roots. Therefore, the roots are real and distinct.