91Ó°ÊÓ

When dealing with quadratic inequalities, the discriminant is a helpful tool to determine the nature of the roots. This is a value that stems from the quadratic equation formula: \( ax^2 + bx + c = 0 \). The discriminant \( \Delta \) is calculated with \( \Delta = b^2 - 4ac \). Understanding the nature of the discriminant can provide insights into the type of roots a quadratic equation possesses:

• If \( \Delta > 0 \), the equation has two distinct real roots. This means the parabola cuts the \( x \)-axis at two points.
• If \( \Delta = 0 \), there is exactly one real root (a double root), where the parabola touches the \( x \)-axis at one point.
• If \( \Delta < 0 \), the equation has no real roots, which means the parabola does not intersect the \( x \)-axis.

In our problem, with \( \Delta = 256 \), we found that \( \Delta > 0 \), indicating two distinct real roots. These roots help to break the number line into segments to test where the inequality holds.

The quadratic formula is a reliable method to find the roots of any quadratic equation \( ax^2 + bx + c = 0 \). This formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]It's designed to compute roots using the coefficients \( a, b, \) and \( c \) from the quadratic equation. Let's break down the components:

• \( -b \pm \sqrt{b^2 - 4ac} \): This part involves the value of the discriminant, helping to determine the nature and number of the roots.
• \( 2a \): Divides the whole expression to correctly determine each root's position on the \( x \)-axis.

Using the quadratic formula in the original exercise, we discovered roots \( x = \frac{3}{2} \) and \( x = \frac{1}{6} \). These solutions represent points where the quadratic equation equals zero. Consequently, they serve as critical points to evaluate where the inequality is satisfied.

Once the roots of the quadratic equation have been determined, interval testing becomes pivotal in solving the inequality. With the quadratic \( 12x^2 - 20x + 3 \), roots \( x = \frac{3}{2} \) and \( x = \frac{1}{6} \) divide the \( x \)-axis into distinct intervals: \((-\infty, \frac{1}{6}), (\frac{1}{6}, \frac{3}{2}), (\frac{3}{2}, \infty)\). Here's how interval testing works:

• Select a test point from each interval. This involves checking points between and beyond the roots to see if they satisfy the inequality.
• Substitute these points back into the quadratic inequality and see if the resultant expression is true.
• If true, then the entire interval satisfies the inequality.
• Don’t forget to account for boundaries, particularly since this is a "greater than or equal" inequality, \( \geq 0 \).

From our testing:- The interval \((-\infty, \frac{1}{6}]\) holds as true since a test at \( x = 0 \) gives \( 3 \geq 0 \).- The interval \((\frac{3}{2}, \infty)\) also holds true as a test at \( x = 2 \) gives \( 11 \geq 0 \).These findings confirm that the solution to the inequality is \( x \in (-\infty, \frac{1}{6}] \cup [\frac{3}{2}, \infty) \), cataloging regions that uphold the original conditions.