# Given one inequality, prove second inequality

I'm stuck on this homework problem. Can someone please give me a hint?

Given $2a^2+3ab+4ac<0$. Is the following expression zero, negative or positive? $$9b^2-32ac$$

Thanks.

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You can remove $a$ from the first inequality. – Damian Sobota Apr 24 '13 at 0:47
Is it possible you mean $2a^2+3ab+4c<0$ and $9b^2-32c$? – Michael Grant Apr 24 '13 at 0:49
@Angie: Take a look at the polynomial $2ax^2 + 3bx + 4c$; if you compute it out, you'll notice that its discriminant is exactly $9b^2 - 32ac$. Now, the important thing about the discriminant is that if the discriminant is negative, the polynomial never crosses the x-axis; if it is exactly zero, then it only touches the x-axis once; and if it is positive, then the polynomial touches the x-axis twice.... – Chris Apr 24 '13 at 1:21
...Now, if $P(x)$ is our polynomial, then $a*P(1)$ is exactly $2a^2 + 3ab + 4ac$. Suppose $a < 0$; then it must be true that $P(1) > 0$. Since $a<0$, the polynomial opens downwards; since $P(x)$ was positive at $x=1$, the polynomial must therefore have two roots. This says that the discriminant must be positive. You should verify that, if $a > 0$, a similar argument shows that the discriminant is positive then, too. Since we cannot have $a = 0$ and $2a^2 + 3ab + 4ac < 0$, this says that the discriminant must always be positive. This gives you your answer! – Chris Apr 24 '13 at 1:28

$$9b^2 - 32ac > 9b^2 + 8(2a^2+3ab) = ? \ge 0$$
Here's my suggestion: Consider the quadratic function $f(x) = 2a^2 x^2 + 3abx + 4ac$. Then your initial inequality can be expressed as $f(1) < 0$, and the discriminant of $f(x)$ is given by $a^2(9b^2 - 32ac)$. Now investigate the behavior of the roots of $f(x)$. It might be useful to observe that $f(x)$ is a "happy" parabola.