Question
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Solve without a calculator. Show your factoring and a complete number-line on your worksheet along with the required information and shading or rough graph with shading. Write your solution using interval notation. \( x^{3}<8 x^{2} \) Solution in interval notation:

Ask by Boyd Rowe. in the United States
Mar 10,2025

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Answer

The solution to the inequality \(x^{3}<8x^{2}\) is \(x \in (-\infty,0) \cup (0,8)\).

Solution

Solve the equation \( x^{3}<8x^{2} \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{3}<8x^{2}\) - step1: Move the expression to the left side: \(x^{3}-8x^{2}<0\) - step2: Rewrite the expression: \(x^{3}-8x^{2}=0\) - step3: Factor the expression: \(x^{2}\left(x-8\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x^{2}=0\\&x-8=0\end{align}\) - step5: Set the base equal to 0: \(\begin{align}&x=0\\&x=8\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<0\\&08\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-1\\&x_{2}=4\\&x_{3}=9\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<0\textrm{ }\textrm{is the solution}\\&08\textrm{ }\textrm{is not a solution}\end{align}\) - step9: The final solution is \(x \in \left(-\infty,0\right)\cup \left(0,8\right):\) \(x \in \left(-\infty,0\right)\cup \left(0,8\right)\) The solution to the inequality \(x^{3}<8x^{2}\) in interval notation is \(x \in (-\infty,0) \cup (0,8)\).

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To solve the inequality \( x^{3} < 8x^{2} \), let's start by rearranging it into standard form: 1. Bring all terms to one side: \[ x^{3} - 8x^{2} < 0 \] 2. Factor out the common term: \[ x^{2}(x - 8) < 0 \] Now we can find the critical points by setting \( x^{2} = 0 \) or \( x - 8 = 0 \): - From \( x^{2} = 0 \): \( x = 0 \) - From \( x - 8 = 0 \): \( x = 8 \) Critical points are at \( x = 0 \) and \( x = 8 \). These points divide the number line into three intervals: \( (-\infty, 0) \), \( (0, 8) \), and \( (8, \infty) \). Now, we’ll test each interval to see where the inequality holds true: 1. **Interval \( (-\infty, 0) \)**: Choose \( x = -1 \) \[ (-1)^{2}(-1 - 8) = 1 \cdot (-9) < 0 \quad \text{(True)} \] 2. **Interval \( (0, 8) \)**: Choose \( x = 1 \) \[ (1)^{2}(1 - 8) = 1 \cdot (-7) < 0 \quad \text{(True)} \] 3. **Interval \( (8, \infty) \)**: Choose \( x = 9 \) \[ (9)^{2}(9 - 8) = 81 \cdot 1 > 0 \quad \text{(False)} \] The inequality holds for the intervals \( (-\infty, 0) \) and \( (0, 8) \). Now, let’s put this all together in interval notation: The solution is: \[ (-\infty, 0) \cup (0, 8) \] On the number line, you'd shade the regions corresponding to these intervals and leave points at \( x = 0 \) and \( x = 8 \) unshaded since these points do not satisfy the strict inequality.

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