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.

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Your answer is incorrect...when x=2 the LHS is 0. –  fretty Jun 9 '12 at 14:20
sorry i typed it wrong –  mgh Jun 9 '12 at 14:22
@Meg Next time try to use LaTeX to type mathematics in this site, or at least leave plenty space: is it 9x or (-9x) what you wrote? Below you've answers for both cases... –  DonAntonio Jun 9 '12 at 14:27

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 $2 < x < 7$.

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$$-9x-14-x^2>0\Longleftrightarrow x^2+9x+14<0$$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 $$(x+7)(x+2)<0\Longrightarrow -7<x<-2$$

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It's $9x$, I think. –  Gigili Jun 9 '12 at 14:27
I think it is $\,-9x\,$, but it's a matter of opinion until the OP addresses the issue. –  DonAntonio Jun 9 '12 at 14:30
hey all its 9x...sorry for the confusion –  mgh Jun 9 '12 at 14:33
@Meg Then Gigili's answer's on the money, but you can take any answer's method and do it yourself...and it'd be nice if you pick up some answer and accept it. –  DonAntonio Jun 9 '12 at 14:39

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 ($x_{i},x_{i+1}$).

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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:-

1. Re-arranging terms, we get $x^2 - 9x + 14 < 0$ (as shown in a previous answer).
2. Solve $x^2 - 9x + 14 = 0$ to get 2 and 7 as the roots.
3. Mark (2, 0) & (7, 0) on the x-axis. They are points that $y = x^2 - 9x + 14$ must go through.
4. 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.
5. Draw the graph according to the results in (4) & (5). See Fig 1.
6. $y = x^2 - 9x + 14$ and $x^2 - 9x + 14 < 0$ together mean $y < 0$.
7. Look for what xs’ will make (6) to happen. Thus, found all the xs are lying between 2 and 7.
8. 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.

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