Solving a quadratic Inequality

Question

My question is:

Solve $$9x-14-x^2>0$$

My answer is: $2 < x < 7$

Though I know my answer is right, I want to know in what ways I can solve it and how it can be graphically represented. Thank you.

Answer 1

Let's rearrange the inequality to get:

$x^2 - 9x + 14 < 0$

i.e.

$(x-7)(x-2) < 0$.

Now the product of two numbers is negative if and only if exactly one of the numbers is negative.

So we have that the inequality is satisfied whenever $x-7 < 0$ or $x-2 < 0$ but not both.

This happens when $2 < x < 7$.

Answer 2

$$-9x-14-x^2>0\Longleftrightarrow x^2+9x+14<0$$Since $\,\,\Delta=81-4\cdot 14=25\,$ , the roots of LHS are $\,\,\displaystyle{x_{1,2}=\frac{-9\pm 5}{2}=-7\,,\,-2}\,$, so the inequality is $$(x+7)(x+2)<0\Longrightarrow -7<x<-2$$

Answer 3

The general way is to use the formula $$ ax^2 + bx + c = 0 \Rightarrow x_{1,2} = \frac{-b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$ and see when the function equals zero. Since it is continuous, it is enough to check the sign of $f(x)$ for some $x_1 < x < x_2$. If it is positve, the $f(x) > 0$ for $x_1 < x < x_2$ and $f(x) < 0$ otherwise. Similarly, if it is negative, the $f(x) < 0$ for $x_1 < x < x_2$ and $f(x) > 0$ otherwise.

The key is that a continuous function has the same sign between two zeroes, so the general way is to find all the zeroes (the $x_i$ where $f(x_i)=0$) and then check by substitution what is its sign between each adjacent pair of them ($ x_{i},x_{i+1} $).