Polynomial Inequalities

This wiki page considers the case where the polynomial involves terms of degree 2 or more. See linear inequalities for the case of degree 1.

A polynomial inequality is an inequality where both sides of the inequality are polynomials. For example, \(x^3 \ge x^4\) is a polynomial inequality which is satisfied if and only if \(0 \le x \le 1.\)

These inequalities can give insight into the behavior of polynomials. For example, because \( x^2 + 2x + 1 = (x+1)^2 \ge 0,\) the minimum possible value of \(x^2 + 2x + 12 = (x^2 + 2x + 1) + 11\) is \(0+11 = 11.\)

This technique also has very practical implications. For example, economic models of revenue \(\big(R(x)\big)\) and costs \(\big(C(x)\big)\) often model these values as polynomials in terms of widgets produced (\(x\)), so being able to solve \(R(x) \ge C(x)\) would tell a business where they are profitable.

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 \(x^2-5x+6>0.\)

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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 \(|x|-2 \sqrt{x} +1 <0.\)

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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 \(x^2 + \frac{1}{x^2} \geq 2.\)

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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:

Most of us know how to solve equations. But what if you were presented an inequality of the form \(\frac{f(x)}{g(x)}\) which is greater than, greater than or equal to, less than, or less than or equal to 0? There are multiple ways to solve this. One way is the wavy curve method.

Solve \(\dfrac{3x - x^2}{{(x + 4)}^2} \geq 0.\)

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Step 1.
Factorize the polynomials: \[\begin{align} \dfrac{x(3 - x)}{{(x + 4)}^2} & \geq 0 \end{align}.\]

Step 2.
Make the coefficient of the variable of all factors positive: \[\begin{align} \dfrac{-x(x - 3)}{{(x + 4)}^2} & \geq 0 \end{align}.\]

Step 3.
Multiply/divide by -1 both sides of the inequality to remove the minus sign (but in doing so, the inequality would reverse): \[\begin{align} \dfrac{x(x - 3)}{{(x + 4)}^2} & \leq 0 \end{align}.\]

Step 4.
Find the roots of the inequality by equating each factor to 0: \[\begin{align} x & = 0\\ x - 3 = 0 \implies x & = 3\\ x + 4 = 0 \implies x & = -4. \end{align}\]

Step 5.
Plot the points on the number line. Now, start with the largest factor, i.e. 3. Initially, a curve from the positive region of the number line should intersect that point (here 3). Now, look at the power of the respective factors. If it is odd, then we have to change the path of the curve from their respective roots. If it is even, continue in the same region. Here, the curve would change its path at 0 and 3 because their factors are odd powers. However, at -4, it would not change its direction since its factor has an even power.

Now, if the inequality is either \(\geq\) or \(\leq\) 0, then we have to consider those values of \(x\) at which the inequality is equal to 0. However, as a rule of the wavy curve method, we should exclude the root of the factor in the denominator (here -4) in our solution set.

So, our final answer is \(x \in [0,3]. \ _\square\)

If \(x\) is a real number, what is the minimum value of \(x^2 - 2x + 123 \)?

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By completing the square, we see that \[x^2-2x + 123 = (x^2 -2 x + 1) + 122 = (x-1)^2 + 122 \geq 0 + 122 = 122.\] So the minimum value of the expression above is \(122\). \(_\square\)

Prove that \(x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 \geq 0 \) for all real \(x\).

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By Pascal's triangle, we have \((x-1)^6 \geq 0 \). To be continued.