How to solve quadratic inequalities with two boundaries

Question

I was solving a math question on an online test, and the question was asking what is possible values of $x$ if the perimeter of some rectangle tile is between $20$ cm and $54$ cm. I wondered how we would find possible values of $x$ if the area was between $20$ and $54$ cm.

I thought we could use sign analysis for each inequality, but I understood that won't work because we will be having different inequalities in each part and that doesn't make sense.

Inequality looks like this: $20<2x^2+12x<54$

How could we solve quadratic inequalities with two boundaries like this?

Answer 1

The values of $x$ that verify the inequality, will verify the following inequalities at the same time: $$ \begin{array} 020<2x^2+12x, & 2x^2+12x<54\end{array} $$ So all we have to do is work with both and we will intersect the solution spaces, getting the solution of the original inequality. Let's start with the first one: $$ 20<2x^2+12x \implies 0<(x+3+\sqrt{19}) ·(x+3-\sqrt{19}) \implies x>-3+\sqrt{19} \ \lor \ x<-3-\sqrt{19} $$ The second one will have the following result: $$ 2x^2+12x<54 \implies x^2+6x-27<0 \implies -9<x<3 $$ The intersection is all of the $x\in \mathbb{R}\ $ such that: $$ -9<x<-3-\sqrt{19} \ \lor -3+\sqrt{19}<x<3 $$

Answer 2

You can solve it by splitting it up into two inequalities, and solving them each separately. If you start from $20<2x^2+12x<54$, that means you need both of the following to be true:

1. $2x^2+12x-20 > 0$
2. $2x^2+12x-54 < 0$

The first quadratic inequality is true for $x<-3-\sqrt{19}$ or for $x > -3+\sqrt{19}$. The second quadratic inequality is true for $-9<x<3$. You can get to these solutions by graphing the quadratic or using the quadratic formula, then checking signs, as you mention in your question.

Combining everything, we need:

$$\left( x<-3-\sqrt{19} \ \cup \ x > -3+\sqrt{19} \right) \ \cap \ (-9<x<3) , $$

where $\cup$ means "or" and $\cap$ means "and."

Then, $-3-\sqrt{19}\approx -7.4$, and $-3+\sqrt{19} \approx 1.4$. So, we end up at the following set of values:

$$ -9<x<-3-\sqrt{19} \ \cup \ -3+\sqrt{19}<x<3. $$

If that last step doesn't make sense and it's hard to visualize, try plotting both of the inequalities on a number line, and wherever they're both true, that's the solution.

Answer 3

Hint:

You just have to solve the system of quadratic inequalities: \begin{cases} 2x^2+12x-20 >0,\\[0.5ex] 2x^2+12x-54<0. \end{cases} The first quadratic polynomial has real roots and its leading coefficient is positive, hence the first inequality is satisfied outside of the interval of the roots.

The second quadratic polynomial does have real roots, hence it has negative values in the interval of the roots.