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\begin{tabular}{|l|l|}\hline 1.1. & \( x^{2}+x-6>0 \) \\ \hline 1.2. & \( -x^{2}+x+2>0 \) \\ \hline 1.3. & \( -x^{2}+x+2 \geq x^{2}+x-6 \) \\ \hline 1.4. & \( x^{2}+x-6 \leq 0 \) \\ \hline\end{tabular}

Ask by Conner Hammond. in South Africa
Feb 20,2025

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Answer

Here are the solutions for each inequality: 1. **1.1.** \( x^{2}+x-6>0 \): \( x < -3 \) or \( x > 2 \) 2. **1.2.** \( -x^{2}+x+2>0 \): \( -1 < x < 2 \) 3. **1.3.** \( -x^{2}+x+2 \geq x^{2}+x-6 \): \( -2 \leq x \leq 2 \) 4. **1.4.** \( x^{2}+x-6 \leq 0 \): \( -3 \leq x \leq 2 \)

Solution

Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}+x-6\leq 0\) - step1: Rewrite the expression: \(x^{2}+x-6=0\) - step2: Factor the expression: \(\left(x-2\right)\left(x+3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-2=0\\&x+3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=2\\&x=-3\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-3\\&-32\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=-1\\&x_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-3\textrm{ }\textrm{is not a solution}\\&-32\textrm{ }\textrm{is not a solution}\end{align}\) - step8: Include the critical value: \(\begin{align}&-3\leq x\leq 2\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(-3\leq x\leq 2:\) \(-3\leq x\leq 2\) Solve the equation \( -x^{2}+x+2>0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-x^{2}+x+2>0\) - step1: Rewrite the expression: \(-x^{2}+x+2=0\) - step2: Factor the expression: \(\left(-x+2\right)\left(x+1\right)=0\) - step3: Separate into possible cases: \(\begin{align}&-x+2=0\\&x+1=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=2\\&x=-1\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-1\\&-12\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=1\\&x_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-1\textrm{ }\textrm{is not a solution}\\&-12\textrm{ }\textrm{is not a solution}\end{align}\) - step8: The final solution is \(-10 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}+x-6>0\) - step1: Rewrite the expression: \(x^{2}+x-6=0\) - step2: Factor the expression: \(\left(x-2\right)\left(x+3\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-2=0\\&x+3=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=2\\&x=-3\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-3\\&-32\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-4\\&x_{2}=-1\\&x_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-3\textrm{ }\textrm{is the solution}\\&-32\textrm{ }\textrm{is the solution}\end{align}\) - step8: The final solution is \(x \in \left(-\infty,-3\right)\cup \left(2,+\infty\right):\) \(x \in \left(-\infty,-3\right)\cup \left(2,+\infty\right)\) Solve the equation \( -x^{2}+x+2 \geq x^{2}+x-6 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-x^{2}+x+2\geq x^{2}+x-6\) - step1: Cancel equal terms: \(-x^{2}+2\geq x^{2}-6\) - step2: Move the expression to the left side: \(-x^{2}+2-\left(x^{2}-6\right)\geq 0\) - step3: Calculate: \(-2x^{2}+8\geq 0\) - step4: Rewrite the expression: \(-2x^{2}+8=0\) - step5: Move the constant to the right side: \(-2x^{2}=0-8\) - step6: Remove 0: \(-2x^{2}=-8\) - step7: Change the signs: \(2x^{2}=8\) - step8: Divide both sides: \(\frac{2x^{2}}{2}=\frac{8}{2}\) - step9: Divide the numbers: \(x^{2}=4\) - step10: Simplify the expression: \(x=\pm \sqrt{4}\) - step11: Simplify: \(x=\pm 2\) - step12: Separate into possible cases: \(\begin{align}&x=2\\&x=-2\end{align}\) - step13: Determine the test intervals: \(\begin{align}&x<-2\\&-22\end{align}\) - step14: Choose a value: \(\begin{align}&x_{1}=-3\\&x_{2}=0\\&x_{3}=3\end{align}\) - step15: Test the chosen value: \(\begin{align}&x<-2\textrm{ }\textrm{is not a solution}\\&-22\textrm{ }\textrm{is not a solution}\end{align}\) - step16: Include the critical value: \(\begin{align}&-2\leq x\leq 2\textrm{ }\textrm{is the solution}\end{align}\) - step17: The final solution is \(-2\leq x\leq 2:\) \(-2\leq x\leq 2\) Here are the solutions for each of the inequalities provided: ### 1.1. Solve \( x^{2}+x-6>0 \) To solve the inequality \( x^{2}+x-6>0 \), we first find the roots of the corresponding equation \( x^{2}+x-6=0 \). The roots are \( x = -3 \) and \( x = 2 \). The solution to the inequality is: \[ x \in (-\infty, -3) \cup (2, +\infty) \] ### 1.2. Solve \( -x^{2}+x+2>0 \) For the inequality \( -x^{2}+x+2>0 \), we find the roots of the equation \( -x^{2}+x+2=0 \). The roots are \( x = -1 \) and \( x = 2 \). The solution to the inequality is: \[ -1 < x < 2 \] ### 1.3. Solve \( -x^{2}+x+2 \geq x^{2}+x-6 \) Rearranging the inequality gives us: \[ -x^{2}+x+2 - (x^{2}+x-6) \geq 0 \] This simplifies to: \[ -2x^{2} + 8 \geq 0 \] Factoring gives: \[ -2(x^{2}-4) \geq 0 \implies -2(x-2)(x+2) \geq 0 \] The roots are \( x = -2 \) and \( x = 2 \). The solution to the inequality is: \[ -2 \leq x \leq 2 \] ### 1.4. Solve \( x^{2}+x-6 \leq 0 \) This is similar to the first inequality. The roots are \( x = -3 \) and \( x = 2 \). The solution to the inequality is: \[ -3 \leq x \leq 2 \] ### Summary of Solutions: - **1.1:** \( x \in (-\infty, -3) \cup (2, +\infty) \) - **1.2:** \( -1 < x < 2 \) - **1.3:** \( -2 \leq x \leq 2 \) - **1.4:** \( -3 \leq x \leq 2 \)

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To solve the inequalities, it helps to factor or analyze the quadratic expressions. For \( x^{2}+x-6>0 \), you can factor it to \( (x-2)(x+3) > 0 \). The critical points are \( x = -3 \) and \( x = 2 \). Testing intervals around these points will reveal that the solution is \( x < -3 \) or \( x > 2 \). Similarly, for \( -x^{2}+x+2>0 \), rearranging gives \( x^{2} - x - 2 < 0 \), factorable to \( (x-2)(x+1) < 0 \). The solution is \( -1 < x < 2 \) after testing intervals. When comparing the two expressions in 1.3, combine them into a single inequality to find the range where one is greater than or equal to the other, facilitating more nuanced analysis of their roots. The final one, \( x^{2}+x-6 \leq 0 \), focuses on the same critical points, where the quadratic is less than or equal to zero between the roots, which leads us back to \( -3 \leq x \leq 2 \).

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