Problem 25 - Algebra and Trigonometry

Question

Solve each polynomial inequality in Exercises \(1-42\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ x^{2} \leq 4 x-2 $$

Step-by-Step Solution

Answer

The solution of the inequality \(x^{2} \leq 4x - 2\) in interval notation is an empty set or \([]\). There are no real numbers which make the inequality true.

Step 1: Rewrite the Inequality

Rewrite the inequality in a standard form. This can be done by subtracting \(4x\) and adding \(2\) on both sides so that all terms are in one side of the inequality: \[x^{2} - 4x + 2 \leq 0\]

Step 2: Factor the Quadratic

Factor the quadratic equation. Unfortunately, it's not possible to factor the polynomial \(x^{2} - 4x + 2\) easily. Thus, we'll use the quadratic formula \(x = [-b \pm \sqrt{b^2-4ac}]/2a\) where \(a=1\), \(b=-4\) and \(c=2\). This yields two roots \(x = [4 \pm \sqrt{(-4)^2-4*1*2}]/2*1 = 2 \pm \sqrt{4-8} = 2 \pm \sqrt{-4} \). The roots are not real numbers as they are complex numbers. This tells us that the quadratic function does not touch or cross the x-axis.

Step 3: Determine the intervals and test the inequality

Since the roots are not real, the graph of this function does not touch or cross the x-axis. It only opens upwards because the coefficient of \(x^2\) is positive. This means all values of \(x\) will result in the expression being positive, except when the inequality reach a minimum value. However, we are looking for values where the quadratic function is less than or equal to zero. Thus, none of the real numbers is a solution. If the minimum value of the graph of the quadratic function is equal or less than zero, all x for that minimum value are solutions.

Step 4: Express the solution in interval notation

There are no real solutions for this inequality, therefore there are no intervals of real numbers where the inequality is true. The solution set in interval notation is expressed as empty set or \(\varnothing \) or \([]\).