Polynomial Inequalities | Brilliant Math & Science Wiki

To solve inequalities involving the quadratic form \(ax^2+bx+c\), we need to consider the basic tools. These tools will help to solve the quadratic inequality problems:

\(\text{1. }ab > 0 \Leftrightarrow a>0\) and \(b>0\) or \(a<0\) and \(b<0\)

\(\text{2. }ab \geq 0 \Leftrightarrow a \geq 0\) and \( b \geq0\) or \(a \leq 0\) and \(b \leq0\)

\(\text{3. }ab < 0 \Leftrightarrow a>0\) and \(b<0\) or \(a<0\) and \(b>0\)

\(\text{4. }ab \leq 0 \Leftrightarrow a \geq 0\) and \(b \leq 0\) or \(a \leq 0\) and \(b \geq0\)

\(\text{5. }a^2 > 0\) when \( a \in \mathbb{R}, a \neq 0\)

\(\text{6. }a^2 \geq 0 \Leftrightarrow \) identity \( \forall a \in \mathbb{R}\)

\(\text{7. }a^2 < 0 \text{ false}\) when \(a \in \mathbb{R}\)

\(\text{8. }a^2 \leq 0 \Leftrightarrow a = 0\)

\(\text{9. }\frac{a}{b} \leq 0 \Leftrightarrow ab \leq 0\) when \(b \neq 0\)

\(\text{10. }\frac{a}{b} < 0 \Leftrightarrow ab < 0\) when \(b \neq 0\)

\(\text{11. }\frac{a}{b} \geq 0 \Leftrightarrow ab \geq 0\) when \(b \neq 0\)

\(\text{12. }\frac{a}{b} > 0 \Leftrightarrow ab > 0\) when \(b \neq 0\)

Here are some examples followed by some problems to try on your own:

Find the solution set of \(x^2-5x+6>0.\)

Since \(x^2-5x+6>0,\) we have \((x-2)(x-3)>0.\) Thus, there are two cases: \(x-2 > 0\) and \(x-3 > 0,\) or \(x-2 < 0\) and \(x-3 < 0.\)

Case 1: \(x-2 > 0 \text{ and }x-3 > 0 \implies x > 2 \text{ and }x>3 \implies x > 3.\) Case 2: \( x < 2 \text{ and } x < 3 \implies x < 2.\)

Therefore, the answer is \(x > 3\) or \(x <2\). \(_\square\)

Find the solution set of \(|x|-2 \sqrt{x} +1 <0.\)

Since \(|x|-2 \sqrt{x} +1 <0,\) we have \(\big( \sqrt{x}-1\big)^2 <0.\) By theorem, \(a^2 < 0 \Leftrightarrow a = \phi.\) Therefore, the answer is \(x \in \phi.\ _\square\)

Find the solution set of \(x^2 + \frac{1}{x^2} \geq 2.\)

Since \(x^2 + \frac{1}{x^2} \geq 2,\) we have \(x^2 -2 + \frac{1}{x^2} \geq 0 \implies \left(x- \frac{1}{x}\right)^2 \geq 0.\) By theorem, \(a^2 \geq 0 \Leftrightarrow\) identity. Therefore, the answer is \(x \in \mathbb{R}\, (x\ne 0).\ _\square\)

Try the following problems:

No range exists \[x < -4 , -3\le x\le 3, x > 4\] None if these choices \[x = 3\] \[ x\le -4 , -3\le x\le 3, x\ge 4 \] \[x \ge 3\] Reveal the answer

\[ \frac { { x }^{ 2 }-9 }{ { x }^{ 2 }-16 } \ge 0\]

Solve the quadratic inequality above.

The correct answer is: \[x < -4 , -3\le x\le 3, x > 4\]

0 18 5 13 12 9 6 Reveal the answer

If \(x\) is an integer that satisfies the inequality

\[ 9 < x ^2 < 99, \]

find the difference between the maximum and minimum possible values of \(x.\)

The correct answer is: 18