Question - Graphical Representation of a Rational Inequality Solution
Solution:
To find the solution to the inequality \((3x + 1)^2 > 3(3x + 1)\), we start by expanding and simplifying:
\((3x + 1)(3x + 1) > 3(3x + 1)\)
\(9x^2 + 6x + 1 > 9x + 3\)
\(9x^2 + 6x - 9x + 1 - 3 > 0\)
\(9x^2 - 3x - 2 > 0\)
We then factor the quadratic expression, if possible, or use the quadratic formula to find the roots (the values of \(x\) where \(9x^2 - 3x - 2 = 0\)).
Factoring gives us:
\((3x + 1)(3x - 2) > 0\)
The roots are \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\). The inequality will change signs at each of these points. To determine where the expression is positive, we can test intervals that are defined by these roots. We choose test points from intervals \((-∞, -\frac{1}{3})\), \((- \frac{1}{3}, \frac{2}{3})\), and \((\frac{2}{3}, ∞)\).
Upon testing, we find the inequality is satisfied for \(x < -\frac{1}{3}\) and \(x > \frac{2}{3}\).
On the number line, the solution will be shown with an open circle at \(x = -\frac{1}{3}\) and \(x = \frac{2}{3}\) and shading to the left of \(x = -\frac{1}{3}\) and to the right of \(x = \frac{2}{3}\), because these are not included in the solution set.
Therefore, the correct answer is option D.
Tutor Solution:
Step 1: Expand the square on the left side of the inequality and distribute the 3 on the right side.
\[ (3x + 1)^2 = (3x + 1)(3x + 1) = 9x^2 + 6x + 1\] \[ 3(3x + 1) = 9x + 3 \]Step 2: Set up the inequality with the expanded terms.
\[ 9x^2 + 6x + 1 > 9x + 3 \]Step 3: Move all terms to one side to get a standard form quadratic inequality.
\[ 9x^2 + 6x + 1 - 9x - 3 > 0 \] \[ 9x^2 - 3x - 2 > 0 \]Step 4: Factor the quadratic expression, if possible. In this case, let's try to factor it.
We look for two numbers that multiply to \( 9 \times -2 = -18 \) and add to -3. Those numbers are -6 and 3. However, the quadratic does not factor nicely. As such, we may need to use the quadratic formula to find the roots.Step 5: Find the roots of the quadratic equation using the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our quadratic \( 9x^2 - 3x - 2 \), \( a = 9 \), \( b = -3 \), and \( c = -2 \). \[ x = \frac{3 \pm \sqrt{(-3)^2 - 4(9)(-2)}}{2(9)} \] \[ x = \frac{3 \pm \sqrt{9 + 72}}{18} \] \[ x = \frac{3 \pm \sqrt{81}}{18} \] \[ x = \frac{3 \pm 9}{18} \] So we have two roots: \[ x_1 = \frac{3 + 9}{18} = \frac{12}{18} = \frac{2}{3} \] \[ x_2 = \frac{3 - 9}{18} = \frac{-6}{18} = \frac{-1}{3} \]Step 6: Determine the intervals of the solution by testing points in the inequality.
The roots divide the number line into three intervals: \( (-\infty, -\frac{1}{3}), (-\frac{1}{3}, \frac{2}{3}), (\frac{2}{3}, \infty) \). For \( x < -\frac{1}{3} \), let \( x = -1 \): \[ 9(-1)^2 - 3(-1) - 2 > 0 \] \[ 9 + 3 - 2 > 0 \] \[ 10 > 0 \] (True, so this interval is part of the solution) For \( -\frac{1}{3} < x < \frac{2}{3} \), let \( x = 0 \): \[ 9(0)^2 - 3(0) - 2 > 0 \] \[ -2 > 0 \] (False, so this interval is not part of the solution) For \( x > \frac{2}{3} \), let \( x = 1 \): \[ 9(1)^2 - 3(1) - 2 > 0 \] \[ 9 - 3 - 2 > 0 \] \[ 4 > 0 \] (True, so this interval is part of the solution)Step 7: Combine the solution intervals and express the answer graphically.
The solution of the inequality \( 9x^2 - 3x - 2 > 0 \) is \( x \in (-\infty, -\frac{1}{3}) \cup (\frac{2}{3}, \infty) \). Since the original inequality uses a strict inequality \( > \), we use open dots to mark the roots on the number line, indicating these points are not included in the solution. Looking at the options provided in the image, the correct answer should show the intervals \( (-\infty, -\frac{1}{3}) \) and \( (\frac{2}{3}, \infty) \) on the number line with open dots at \( x = -\frac{1}{3} \) and \( x = \frac{2}{3} \). Option D in the image is the correct representation of this solution.CamTutor
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