If c 0 and the equation 3ax^2 + 4bx + c = 0 | vedclass

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(C) Let $f(x) = 3ax^2 + 4bx + c$.Since the equation $f(x) = 0$ has no real roots and $c > 0$ (which is $f(0) > 0$),the parabola $y = f(x)$ must lie en

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$3a + c > 4b$

Vedclass · Answer

(C) Let $f(x) = 3ax^2 + 4bx + c$.Since the equation $f(x) = 0$ has no real roots and $c > 0$ (which is $f(0) > 0$),the parabola $y = f(x)$ must lie entirely above the $x$-axis.This implies that $f(x) > 0$ for all real values of $x$.Specifically,for $x = -1$,we have $f(-1) > 0$.Substituting $x = -1$ into the equation,we get $f(-1) = 3a(-1)^2 + 4b(-1) + c = 3a - 4b + c$.Since $f(-1) > 0$,it follows that $3a - 4b + c > 0$.Rearranging the terms,we get $3a + c > 4b$.

If $c > 0$ and the equation $3ax^2 + 4bx + c = 0$ has no real root,then :-

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