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,the graph of $f(x)$ does not intersect the $x$-axis.Given $c > 0$,we

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$3a + c > 4b$

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(C) Let $f(x) = 3ax^2 + 4bx + c$.Since the equation $f(x) = 0$ has no real roots,the graph of $f(x)$ does not intersect the $x$-axis.Given $c > 0$,we have $f(0) = c > 0$. Since $f(0) > 0$ and there are no real roots,the parabola must open upwards,meaning $3a > 0$,so $a > 0$.Thus,$f(x) > 0$ for all real $x$.Specifically,$f(-1) > 0$.Substituting $x = -1$ into the expression,we get $f(-1) = 3a(-1)^2 + 4b(-1) + c = 3a - 4b + c$.Since $f(-1) > 0$,we have $3a - 4b + c > 0$,which implies $3a + c > 4b$.

Quadratic expression	The minimum value
i) $x^2 + 4x + 6$	a) $1$
ii) $x^2 - 2x + 5$	b) $2$
iii) $x^2 + 6x + 18$	c) $4$
iv) $x^2 - 4x + 5$	d) $9$

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

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