B. Exercises solve and graph the solution set. \( \begin{array}{lll}\text { 15. } x^{2}<4 x+45 & \text { 16. } 6-x^{2} \leq x & \text { 17. } 3 x-1>3-x^{2} \\ \text { 18. } 16 x^{2}-9 \leq 0 & \text { 19. } x+6 \leq 2 x^{2} & \text { 20. } x^{2}-6 x+7>0 \\ \text { 21. } 12 x+3 \leq 4 x^{2} & \text { 22. } x-x^{2}<1 & \text { 23. } x^{2}+2 x>-1\end{array} \)

Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}+2x+1>0\) - step1: Rewrite the expression: \(x^{2}+2x+1=0\) - step2: Factor the expression: \(\left(x+1\right)^{2}=0\) - step3: Simplify the expression: \(x+1=0\) - step4: Move the constant to the right side: \(x=0-1\) - step5: Remove 0: \(x=-1\) - step6: Determine the test intervals: \(\begin{align}&x<-1\\&x>-1\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=0\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-1\textrm{ }\textrm{is the solution}\\&x>-1\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(x\neq -1:\) \(x\neq -1\) Solve the equation \( x-x^{2}-1<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x-x^{2}-1<0\) - step1: Rewrite the expression: \(x-x^{2}-1=0\) - step2: Add or subtract both sides: \(x-x^{2}=1\) - step3: Divide both sides: \(\frac{x-x^{2}}{-1}=\frac{1}{-1}\) - step4: Evaluate: \(-x+x^{2}=-1\) - step5: Add the same value to both sides: \(-x+x^{2}+\frac{1}{4}=-1+\frac{1}{4}\) - step6: Simplify the expression: \(\left(x-\frac{1}{2}\right)^{2}=-\frac{3}{4}\) - step7: The statement is false for any value of \(x:\) \(x \notin \mathbb{R}\) - step8: Choose a value: \(x=0\) - step9: Test the chosen value: \(x \in \mathbb{R}\) Solve the equation \( 16x^{2}-9\leq0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(16x^{2}-9\leq 0\) - step1: Rewrite the expression: \(16x^{2}-9=0\) - step2: Move the constant to the right side: \(16x^{2}=0+9\) - step3: Remove 0: \(16x^{2}=9\) - step4: Divide both sides: \(\frac{16x^{2}}{16}=\frac{9}{16}\) - step5: Divide the numbers: \(x^{2}=\frac{9}{16}\) - step6: Simplify the expression: \(x=\pm \sqrt{\frac{9}{16}}\) - step7: Simplify: \(x=\pm \frac{3}{4}\) - step8: Separate into possible cases: \(\begin{align}&x=\frac{3}{4}\\&x=-\frac{3}{4}\end{align}\) - step9: Determine the test intervals: \(\begin{align}&x<-\frac{3}{4}\\&-\frac{3}{4}\frac{3}{4}\end{align}\) - step10: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=0\\&x_{3}=2\end{align}\) - step11: Test the chosen value: \(\begin{align}&x<-\frac{3}{4}\textrm{ }\textrm{is not a solution}\\&-\frac{3}{4}\frac{3}{4}\textrm{ }\textrm{is not a solution}\end{align}\) - step12: Include the critical value: \(\begin{align}&-\frac{3}{4}\leq x\leq \frac{3}{4}\textrm{ }\textrm{is the solution}\end{align}\) - step13: The final solution is \(-\frac{3}{4}\leq x\leq \frac{3}{4}:\) \(-\frac{3}{4}\leq x\leq \frac{3}{4}\) Solve the equation \( 6-x^{2}-x\leq0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(6-x^{2}-x\leq 0\) - step1: Rewrite the expression: \(6-x^{2}-x=0\) - step2: Factor the expression: \(\left(2-x\right)\left(3+x\right)=0\) - step3: Separate into possible cases: \(\begin{align}&2-x=0\\&3+x=0\end{align}\) - 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: Include the critical value: \(\begin{align}&x\leq -3\textrm{ }\textrm{is the solution}\\&x\geq 2\textrm{ }\textrm{is the solution}\end{align}\) - step9: 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 \( 3x-1-3+x^{2}>0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(3x-1-3+x^{2}>0\) - step1: Subtract the numbers: \(3x-4+x^{2}>0\) - step2: Rewrite the expression: \(3x-4+x^{2}=0\) - step3: Factor the expression: \(\left(-1+x\right)\left(4+x\right)=0\) - step4: Separate into possible cases: \(\begin{align}&-1+x=0\\&4+x=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=1\\&x=-4\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<-4\\&-41\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-5\\&x_{2}=-2\\&x_{3}=2\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-4\textrm{ }\textrm{is the solution}\\&-41\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(x \in \left(-\infty,-4\right)\cup \left(1,+\infty\right):\) \(x \in \left(-\infty,-4\right)\cup \left(1,+\infty\right)\) Solve the equation \( x^{2}-4x-45<0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}-4x-45<0\) - step1: Rewrite the expression: \(x^{2}-4x-45=0\) - step2: Factor the expression: \(\left(x-9\right)\left(x+5\right)=0\) - step3: Separate into possible cases: \(\begin{align}&x-9=0\\&x+5=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=9\\&x=-5\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-5\\&-59\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-6\\&x_{2}=2\\&x_{3}=10\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-5\textrm{ }\textrm{is not a solution}\\&-59\textrm{ }\textrm{is not a solution}\end{align}\) - step8: The final solution is \(-50 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x^{2}-6x+7>0\) - step1: Rewrite the expression: \(x^{2}-6x+7=0\) - step2: Add or subtract both sides: \(x^{2}-6x=-7\) - step3: Add the same value to both sides: \(x^{2}-6x+9=-7+9\) - step4: Simplify the expression: \(\left(x-3\right)^{2}=2\) - step5: Simplify the expression: \(x-3=\pm \sqrt{2}\) - step6: Separate into possible cases: \(\begin{align}&x-3=\sqrt{2}\\&x-3=-\sqrt{2}\end{align}\) - step7: Move the constant to the right side: \(\begin{align}&x=\sqrt{2}+3\\&x=-\sqrt{2}+3\end{align}\) - step8: Determine the test intervals: \(\begin{align}&x<-\sqrt{2}+3\\&-\sqrt{2}+3\sqrt{2}+3\end{align}\) - step9: Choose a value: \(\begin{align}&x_{1}=1\\&x_{2}=3\\&x_{3}=5\end{align}\) - step10: Test the chosen value: \(\begin{align}&x<-\sqrt{2}+3\textrm{ }\textrm{is the solution}\\&-\sqrt{2}+3\sqrt{2}+3\textrm{ }\textrm{is the solution}\end{align}\) - step11: The final solution is \(x \in \left(-\infty,-\sqrt{2}+3\right)\cup \left(\sqrt{2}+3,+\infty\right):\) \(x \in \left(-\infty,-\sqrt{2}+3\right)\cup \left(\sqrt{2}+3,+\infty\right)\) Solve the equation \( x+6-2x^{2}\leq0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x+6-2x^{2}\leq 0\) - step1: Rewrite the expression: \(x+6-2x^{2}=0\) - step2: Factor the expression: \(\left(2-x\right)\left(3+2x\right)=0\) - step3: Separate into possible cases: \(\begin{align}&2-x=0\\&3+2x=0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=2\\&x=-\frac{3}{2}\end{align}\) - step5: Determine the test intervals: \(\begin{align}&x<-\frac{3}{2}\\&-\frac{3}{2}2\end{align}\) - step6: Choose a value: \(\begin{align}&x_{1}=-3\\&x_{2}=0\\&x_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&x<-\frac{3}{2}\textrm{ }\textrm{is the solution}\\&-\frac{3}{2}2\textrm{ }\textrm{is the solution}\end{align}\) - step8: Include the critical value: \(\begin{align}&x\leq -\frac{3}{2}\textrm{ }\textrm{is the solution}\\&x\geq 2\textrm{ }\textrm{is the solution}\end{align}\) - step9: The final solution is \(x \in \left(-\infty,-\frac{3}{2}\right]\cup \left[2,+\infty\right):\) \(x \in \left(-\infty,-\frac{3}{2}\right]\cup \left[2,+\infty\right)\) Solve the equation \( 12x+3-4x^{2}\leq0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(12x+3-4x^{2}\leq 0\) - step1: Rewrite the expression: \(12x+3-4x^{2}=0\) - step2: Add or subtract both sides: \(12x-4x^{2}=-3\) - step3: Divide both sides: \(\frac{12x-4x^{2}}{-4}=\frac{-3}{-4}\) - step4: Evaluate: \(-3x+x^{2}=\frac{3}{4}\) - step5: Add the same value to both sides: \(-3x+x^{2}+\frac{9}{4}=\frac{3}{4}+\frac{9}{4}\) - step6: Simplify the expression: \(\left(x-\frac{3}{2}\right)^{2}=3\) - step7: Simplify the expression: \(x-\frac{3}{2}=\pm \sqrt{3}\) - step8: Separate into possible cases: \(\begin{align}&x-\frac{3}{2}=\sqrt{3}\\&x-\frac{3}{2}=-\sqrt{3}\end{align}\) - step9: Solve the equation: \(\begin{align}&x=\frac{2\sqrt{3}+3}{2}\\&x=\frac{-2\sqrt{3}+3}{2}\end{align}\) - step10: Determine the test intervals: \(\begin{align}&x<\frac{-2\sqrt{3}+3}{2}\\&\frac{-2\sqrt{3}+3}{2}\frac{2\sqrt{3}+3}{2}\end{align}\) - step11: Choose a value: \(\begin{align}&x_{1}=-1\\&x_{2}=2\\&x_{3}=4\end{align}\) - step12: Test the chosen value: \(\begin{align}&x<\frac{-2\sqrt{3}+3}{2}\textrm{ }\textrm{is the solution}\\&\frac{-2\sqrt{3}+3}{2}\frac{2\sqrt{3}+3}{2}\textrm{ }\textrm{is the solution}\end{align}\) - step13: Include the critical value: \(\begin{align}&x\leq \frac{-2\sqrt{3}+3}{2}\textrm{ }\textrm{is the solution}\\&x\geq \frac{2\sqrt{3}+3}{2}\textrm{ }\textrm{is the solution}\end{align}\) - step14: The final solution is \(x \in \left(-\infty,\frac{-2\sqrt{3}+3}{2}\right]\cup \left[\frac{2\sqrt{3}+3}{2},+\infty\right):\) \(x \in \left(-\infty,\frac{-2\sqrt{3}+3}{2}\right]\cup \left[\frac{2\sqrt{3}+3}{2},+\infty\right)\) Let's solve and graph the solution sets for each inequality step by step. ### 15. \( x^{2} < 4x + 45 \) Rearranging gives: \[ x^{2} - 4x - 45 < 0 \] Factoring: \[ (x - 9)(x + 5) < 0 \] The critical points are \( x = -5 \) and \( x = 9 \). Testing intervals: - For \( x < -5 \): Choose \( x = -6 \) → Positive - For \( -5 < x < 9 \): Choose \( x = 0 \) → Negative - For \( x > 9 \): Choose \( x = 10 \) → Positive Thus, the solution is: \[ -5 < x < 9 \] ### 16. \( 6 - x^{2} \leq x \) Rearranging gives: \[ -x^{2} - x + 6 \leq 0 \quad \Rightarrow \quad x^{2} + x - 6 \geq 0 \] Factoring: \[ (x - 2)(x + 3) \geq 0 \] The critical points are \( x = -3 \) and \( x = 2 \). Testing intervals: - For \( x < -3 \): Positive - For \( -3 < x < 2 \): Negative - For \( x > 2 \): Positive Thus, the solution is: \[ x \in (-\infty, -3] \cup [2, +\infty) \] ### 17. \( 3x - 1 > 3 - x^{2} \) Rearranging gives: \[ x^{2} + 3x - 4 > 0 \] Factoring: \[ (x - 1)(x + 4) > 0 \] The critical points are \( x = -4 \) and \( x = 1 \). Testing intervals: - For \( x < -4 \): Positive - For \( -4 < x < 1 \): Negative - For \( x > 1 \): Positive Thus, the solution is: \[ x \in (-\infty, -4) \cup (1, +\infty) \] ### 18. \( 16x^{2} - 9 \leq 0 \) Rearranging gives: \[ (4x - 3)(4x + 3) \leq 0 \] The critical points are \( x = -\frac{3}{4} \) and \( x = \frac{3}{4} \). Testing intervals: - For \( x < -\frac{3}{4} \): Positive - For \( -\frac{3}{4} < x < \frac{3}{4} \): Negative - For \( x > \frac{3}{4} \): Positive Thus, the solution is: \[ -\frac{3}{4} \leq x \leq \frac{3}{4} \] ### 19. \( x + 6 \leq 2x^{2} \) Rearranging gives: \[ 2x^{2} - x - 6 \geq 0 \] Factoring: \[ (2x + 3)(x - 2) \geq 0 \] The critical points are \( x = -\frac{3}{2} \) and \( x = 2 \). Testing intervals: - For \( x < -\frac{3}{2} \): Positive - For \( -\frac{3}{2} < x < 2 \): Negative - For \( x > 2 \): Positive Thus, the solution is: \[ x \in (-\infty, -\frac{3}{2}] \cup [2, +\infty) \] ### 20. \( x^{2} - 6x + 7 > 0 \) The discriminant is negative: \[ (-6)^{2} - 4 \cdot 1 \cdot 7 = 36 - 28 = 8 > 0 \] Thus, the quadratic does not cross the x-axis and opens upwards. Therefore, it is always positive: \[ x \in \mathbb{R} \] ### 21. \( 12x + 3 \leq 4x^{2} \) Rearranging gives: \[ 4x^{2} - 12x - 3 \geq 0 \] Using the quadratic formula: \[ x = \frac{12 \pm \sqrt{144 + 48}}{8} = \frac{12 \pm 6\sqrt{3}}{8} = \frac{3 \pm \frac{3\sqrt{3}}{2}}{2} \] Thus, the solution is: \[ x \in (-\infty, \frac{-2\sqrt{3}+3}{2}] \cup [\frac{2\sqrt{3}+3}{2}, +\infty) \] ### 22. \( x - x^{2} < 1 \) Rearranging gives: \[ -x^{2} + x - 1 < 0 \quad \Rightarrow \quad x^{2} - x + 1 > 0 \] The discriminant is negative: \[ (-1)^{2} - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 < 0 \] Thus, the quadratic is always positive: \[ x \in \mathbb{R} \] ### 23. \( x^{2} + 2x > -1 \) Rearranging gives: \[ x^{2} + 2x + 1 > 0 \quad \Rightarrow \quad (x + 1)^{2} > 0 \] This is true for all \( x \neq -1 \): \[ x \neq -1 \] ### Summary of Solutions 1. \( -5 < x < 9 \) 2. \( x \in (-\infty, -3] \cup [2, +\infty) \) 3. \( x \in (-\infty, -4) \cup (1, +\infty) \) 4. \( -\frac{3}{4} \leq x \leq \frac{3}{4} \) 5. \( x \in (-\infty, -\frac{3}{2}] \cup [2, +\infty) \) 6. \( x \in \mathbb{R} \) 7. \( x \in (-\infty, \frac{-2\sqrt{3}+3}{2}] \cup