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Magazine Chapter 11: Problem 22
Solve each inequality, and graph the solution set. $$ 3 x^{2}-6 x+2 \leq 0 $$
Short Answer
Expert verified
The solution set is \( 1-\frac{\sqrt{3}}{3} \leq x \leq 1+\frac{\sqrt{3}}{3} \) and is represented by a thick line segment on the number line.
Step by step solution
01
Understand the inequality
Identify the quadratic inequality to solve: \[ 3x^{2}-6x+2 \leq0 \]
02
Rewrite the quadratic inequality in standard form
Confirm that the inequality is already in the standard form: \[ ax^{2}+bx+c \leq 0 \]. Here, \( a = 3 \), \( b = -6 \), and \( c = 2 \).
03
Find the roots of the quadratic equation
Solve the quadratic equation \( 3x^{2}-6x+2 = 0 \) using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substitute \( a \), \( b \), and \( c \): \[ x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3} \]. So, the roots are \( x = 1 \pm \frac{\sqrt{3}}{3} \).
04
Determine intervals to test
The roots divide the number line into three intervals: \( -\infty, 1-\frac{\sqrt{3}}{3} \), \( 1-\frac{\sqrt{3}}{3} , 1+\frac{\sqrt{3}}{3} \), and \( 1+\frac{\sqrt{3}}{3}, +\infty \).
05
Test each interval
Select test points within each interval to determine where \( 3x^{2}-6x+2 \leq 0 \): For \( -\infty < x < 1-\frac{\sqrt{3}}{3} \), test point \( x = 0 \): \( 3(0)^{2} - 6(0) + 2 = 2 > 0 \). For \( 1-\frac{\sqrt{3}}{3} < x < 1+\frac{\sqrt{3}}{3} \), test point \( x = 1 \): \( 3(1)^{2} - 6(1) + 2 = 3 - 6 + 2 = -1 \leq 0 \). For \( 1+\frac{\sqrt{3}}{3} < x < +\infty \), test point \( x = 2 \): \( 3(2)^{2} - 6(2) + 2 = 12 - 12 + 2 = 2 > 0 \).
06
Combine the solution
Combine the intervals where the inequality holds true. From the test points, the quadratic is non-positive in the interval: \[ 1-\frac{\sqrt{3}}{3} \leq x \leq 1+\frac{\sqrt{3}}{3} \].
07
Graph the solution set
Graph the interval on a number line. Draw a thick line segment from \( 1-\frac{\sqrt{3}}{3} \) to \( 1+\frac{\sqrt{3}}{3} \) including the endpoints.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
The quadratic formula is a powerful tool used to solve quadratic equations, which are in the form of ax^2 + bx + c = 0. It provides the solutions by using the coefficients of the equation. The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. First, identify the coefficients a, b, and c. In our inequality, these are: a = 3, b = -6, and c = 2. Plug these values into the quadratic formula to find the roots: \[ x = \frac{6 \pm \sqrt{36 - 24}}{6} = \frac{6 \pm 2\sqrt{3}}{6} = \frac{3 \pm \sqrt{3}}{3} \]. These roots are the points 1 卤 鈭�3/3. They help us solve the inequality and find the intervals to test later.
inequality graphing
Graphing inequalities helps visualize the solution sets on a number line. For the quadratic inequality, \[ 3x^2 - 6x + 2 \leq 0 \], we use the roots found with the quadratic formula, which divide the x-axis into intervals. These roots are critical points where the curve touches or intersects the x-axis.
interval testing
After identifying the roots of the quadratic equation, we divide the number line into intervals based on these roots. The intervals here are: \[ -\infty, 1-\frac{\sqrt{3}}{3} \], \[ 1-\frac{\sqrt{3}}{3}, 1+\frac{\sqrt{3}}{3} \], and \[ 1+\frac{\sqrt{3}}{3}, +\infty \]. We then test a value within each interval to see if the original inequality holds. For instance:
- For \( -\infty < x < 1-\frac{\sqrt{3}}{3} \), use test point \( x = 0 \): 3(0)^2 - 6(0) + 2 = 2 > 0.
- For \( 1-\frac{\sqrt{3}}{3} < x < 1+\frac{\sqrt{3}}{3} \), use test point \( x = 1 \): 3(1)^2 - 6(1) + 2 = -1 <= 0.
- For \( 1+\frac{\sqrt{3}}{3} < x < +\infty \), use test point \( x = 2 \): 3(2)^2 - 6(2) + 2 = 2 > 0.
solution set
The solution set of an inequality is the range of values that satisfy the inequality. From our interval testing, we found that the inequality \( 3x^2 - 6x + 2 \leq 0 \) holds true in the interval \[ 1-\frac{\sqrt{3}}{3}, 1+\frac{\sqrt{3}}{3} \]. This means the values within this interval, including the endpoints, make the inequality true. To visualize this:
- Draw a number line.
- Mark and include the points \( 1-\frac{\sqrt{3}}{3} \) and \( 1+\frac{\sqrt{3}}{3} \).
- Shade the region between and including these points.
This shaded area represents the set of x-values that solve the inequality.
- Draw a number line.
- Mark and include the points \( 1-\frac{\sqrt{3}}{3} \) and \( 1+\frac{\sqrt{3}}{3} \).
- Shade the region between and including these points.
This shaded area represents the set of x-values that solve the inequality.
