When is $12x^2+7x-10$ negative?

When is $12x^2+7x-10$ negative?

Question

1. Find the range of values of $x$ for which the expression $12x^2+7x-10$ is negative.
2. Find the range of values of $c$ for which the expression $x^2+4x+c$ is always positive.

Suggestion

If a quadratic is always positive, what can be said about its roots?

Solution

This is the same as asking: when is the curve $y = 12x^2+7x-10$ below the $x$-axis?

Let’s start by finding when $y = 0$. We can use the quadratic formula here, giving the roots as $\frac{-7 \pm \sqrt{49 + 480}}{24} = -\dfrac{5}{4}\ \text{and}\ \dfrac{2}{3}.$

(We could factorise here, but the factorisation $(3x-2)(4x+5)$ could take us a while to spot.)

The coefficient of $x^2$ is positive here, so the curve is ‘vertex-down’. We have enough information now to sketch the curve.

So the expression $12x^2+7x-10$ is negative for $-\dfrac{5}{4}<x<\dfrac{2}{3}$.

We note that since the coefficient of $x^2$ is positive, $x^2+4x+c$ will always be positive for some values of $x$, and the curve $y=x^2+4x+c$ is ‘vertex-down’.

So the question is the same as asking: which values of $c$ mean that the curve $y = x^2+4x+c$ is always above the $x$-axis?

This means there are no roots for $y = 0$, and so the discriminant must be negative.

So we have $4^2-4c < 0$, or $c > 4$.

Hence the expression is always positive if and only if $c > 4$.

Alternatively we could complete the square here. The expression $x^2+4x+c$ can be written $(x+2)^2 + c-4$, which is always positive if and only if $c>4$.