If a + b + c = 0,then the roots of the equati | vedclass

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(C) Given the quadratic equation $4ax^2 + 3bx + 2c = 0$. The discriminant $D$ is given by $D = B^2 - 4AC = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac$. Given $a

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(C) Given the quadratic equation $4ax^2 + 3bx + 2c = 0$. The discriminant $D$ is given by $D = B^2 - 4AC = (3b)^2 - 4(4a)(2c) = 9b^2 - 32ac$. Given $a + b + c = 0$,we have $b = -(a + c)$. Substituting $b$ into the discriminant: $D = 9(-(a + c))^2 - 32ac = 9(a^2 + 2ac + c^2) - 32ac = 9a^2 + 18ac + 9c^2 - 32ac = 9a^2 - 14ac + 9c^2$. We can rewrite this as $D = 9a^2 - 14ac + 9c^2 = (3a - 3c)^2 + 4ac$. Alternatively,consider the function $f(x) = 4ax^2 + 3bx + 2c$. Since $a+b+c=0$,$b = -a-c$. $f(1) = 4a + 3b + 2c = 4a + 3(-a-c) + 2c = a - c$. $f(0) = 2c$. $f(-1) = 4a - 3b + 2c = 4a - 3(-a-c) + 2c = 7a + 5c$. Since the quadratic takes values of different signs for different $x$,the roots must be real.

If $a + b + c = 0$,then the roots of the equation $4ax^2 + 3bx + 2c = 0$ are

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