Problem 28 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 32 , find the solution set of the given inequality and illustrate the solution on the real number $$ 2 x^{2}-6 x+3<0 $$

Step-by-Step Solution

Answer

The solution set is \[ (\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) \].

Step 1 - Understand the Inequality

The given inequality is \[ 2x^2 - 6x + 3 < 0 \]. This is a quadratic inequality of the form \[ ax^2 + bx + c < 0 \].

Step 2 - Find the Roots of the Quadratic

To solve the inequality, first find the roots of the quadratic equation \[ 2x^2 - 6x + 3 = 0 \]. \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{(-6)^2 - 4(2)(3)}}{2(2)} = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2} \]. So the roots are \[ x = \frac{3 + \sqrt{3}}{2} \] and \[ x = \frac{3 - \sqrt{3}}{2} \].

Step 3 - Test Intervals Between and Beyond the Roots

Check the intervals determined by the roots. These intervals are \[ (-\infty, \frac{3 - \sqrt{3}}{2}), (\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}), (\frac{3 + \sqrt{3}}{2}, \infty) \]. Choose test points from each interval to determine where the inequality holds.

Step 4 - Evaluate the Inequality in Each Interval

Evaluate the quadratic expression at a test point from each interval. Choose \[ x = 0 \] for the interval \[ (-\infty, \frac{3 - \sqrt{3}}{2}) \], \[ x = 1 \] for the interval \[ (\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) \], and \[ x = 2 \] for the interval \[ (\frac{3 + \sqrt{3}}{2}, \infty) \]. Substituting these points into \[ 2x^2 - 6x + 3 \], we find that it is negative in the middle interval.

Step 5 - Write the Solution Set

The inequality \[ 2x^2 - 6x + 3 < 0 \] holds for \[ x \] values in the interval \[ (\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2}) \].

Step 6 - Illustrate the Solution on the Real Number Line

Plot the roots \[ \frac{3 - \sqrt{3}}{2} \] and \[ \frac{3 + \sqrt{3}}{2} \] on the number line and shade the interval between them to represent the solution.

Key Concepts

solving inequalities

Solving inequalities helps us understand ranges of values where a condition holds true. In our case, we’re dealing with a quadratic inequality, which is a polynomial inequality where the highest exponent of the variable (usually x) is 2. These kinds of inequalities often create regions on the number line where the inequality holds true.

Once we understand this, our goal becomes finding those regions. This process typically involves:

• Finding the roots of the associated quadratic equation.
• Dividing the number line based on these roots.
• Testing these regions to see where the inequality holds.

Using these steps, we break down a complex problem into manageable parts.

quadratic equations

A quadratic equation takes the form \[ax^2 + bx + c = 0\]. Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). Quadratic equations have maximum two real roots, and finding these roots is crucial to solving quadratic inequalities.

Using the quadratic formula, \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], we substitute in the coefficients from our quadratic equation. For example, in the inequality \[2x^2 - 6x + 3 < 0\], \(a = 2\), \(b = -6\), and \(c = 3\). Substituting these values into the quadratic formula, we find the roots to be:
\[x = \frac{3 + \sqrt{3}}{2}\]
and
\[x = \frac{3 - \sqrt{3}}{2}\].
Finding these roots helps us pinpoint critical points which divide the number line into different intervals.

interval testing

Interval testing involves checking whether the inequality holds within specific regions of the number line divided by the roots. This is key to identifying solution sets for inequalities.

After finding roots \[x = \frac{3 + \sqrt{3}}{2}\] and \[x = \frac{3 - \sqrt{3}}{2}\], we determine the intervals:

• \((-\infty, \frac{3 - \sqrt{3}}{2})\)
• \((\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2})\)
• \((\frac{3 + \sqrt{3}}{2}, \infty)\)

We choose test points within these intervals to check the sign of the quadratic expression. It ensures which intervals satisfy the inequality. For example:

• For \(x = 0\) in \((-\infty, \frac{3 - \sqrt{3}}{2})\), substituting gives a positive value, so it doesn't satisfy \(2x^2 - 6x + 3 < 0\).
• For \(x = 1\) in \((\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2})\), substituting gives a negative value, so this interval satisfies the inequality.
• For \(x = 2\) in \((\frac{3 + \sqrt{3}}{2}, \infty)\), substituting gives a positive value, so it doesn't satisfy \(2x^2 - 6x + 3 < 0\).

By testing, we confirm the inequality \(2x^2 - 6x + 3 < 0\) holds for \(x\) in the interval \((\frac{3 - \sqrt{3}}{2}, \frac{3 + \sqrt{3}}{2})\).

roots of quadratic equations

The roots of quadratic equations are the values of \(x\) that make the quadratic expression equal to zero. They can be real or complex. For our inequality \(2x^2 - 6x + 3 < 0\), the roots are used to divide the number line.

Finding the roots involves solving the associated quadratic equation \(2x^2 - 6x + 3 = 0\). Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we plug in \(a = 2\), \(b = -6\), and \(c = 3\) to get:
\[x = \frac{6 \pm \sqrt{36 - 24}}{4} = \frac{6 \pm \sqrt{12}}{4} = \frac{6 \pm 2\sqrt{3}}{4} = \frac{3 \pm \sqrt{3}}{2}\].
The roots we found are:

• \(x = \frac{3 + \sqrt{3}}{2}\)
• \(x = \frac{3 - \sqrt{3}}{2}\)

These roots split the number line into regions, aiding in our interval testing process and ultimately helping to solve the inequality. By understanding the roots, we manage the problem systematically.