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.
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.
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 $2 < x < 7$.
$-9x-14-x^2>0\Longleftrightarrow x^2+9x+14<0$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
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} $).
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:-
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.