Solving a quadratic Inequality 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.
 A: 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 $2 < x < 7$.
A: $$-9x-14-x^2>0\Longleftrightarrow x^2+9x+14<0$$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 $$(x+7)(x+2)<0\Longrightarrow -7<x<-2$$
A: 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} $).
A: Though the question was answered more than 15 months ago, I still found that the part “….. and how it can be graphically represented(?)” was not shown in the form of a graph. The following is the required procedures:- 


*

*Re-arranging terms, we get $x^2 - 9x + 14 < 0$ (as shown in a previous answer).

*Solve $x^2 - 9x + 14 = 0$ to get 2 and 7 as the roots.

*Mark (2, 0) & (7, 0) on the x-axis. They are points that $y = x^2 - 9x + 14$ must go through.

*The coefficient of $x^2$ is positive indicating the graph of $y = x^2 - 9x + 14$ is an U-shaped parabola that opens its mouth upward.

*Draw the graph according to the results in (4) & (5). See Fig 1.

*$y = x^2 - 9x + 14$ and $x^2 - 9x + 14 < 0$ together mean $y < 0$.

*Look for what xs’ will make (6) to happen. Thus, found all the xs are lying between 2 and 7.

*Represent the result found (7) graphically and get Fig. 2.

Note-1. Open circles are used to indicate the endpoint(s) are not included.
Note-2. All these may seem a bit tedious but it can be completed within a minute or two.
