OIISTIGN I 11. Solveforx \[ 111 x^{2}+5 x-14=0 \] 1.2. \( 2 x^{2}-3 x-4=0 \) (correct to TWO decimal places) (3) 1.1.3, \( x(x-4) \geq 5 \) 1.1.4. \( \sqrt{7-2 x}-x-4=0 \)

Solve the quadratic equation by following steps: - step0: Solve by factoring: \(111x^{2}+5x-14=0\) - step1: Factor the expression: \(\left(3x-1\right)\left(37x+14\right)=0\) - step2: Separate into possible cases: \(\begin{align}&3x-1=0\\&37x+14=0\end{align}\) - step3: Solve the equation: \(\begin{align}&x=\frac{1}{3}\\&x=-\frac{14}{37}\end{align}\) - step4: Rewrite: \(x_{1}=-\frac{14}{37},x_{2}=\frac{1}{3}\) Solve the equation \( \sqrt{7-2 x}-x-4=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{7-2x}-x-4=0\) - step1: Find the domain: \(\sqrt{7-2x}-x-4=0,x\leq \frac{7}{2}\) - step2: Move the expression to the right-hand side: \(\sqrt{7-2x}=x+4\) - step3: Evaluate: \(\sqrt{7-2x}=x+4,x+4\geq 0\) - step4: Evaluate: \(\sqrt{7-2x}=x+4,x\geq -4\) - step5: Solve the equation: \(\begin{align}&x=-1\\&x=-9\end{align},x\geq -4\) - step6: Find the intersection: \(x=-1\) - step7: Check if the solution is in the defined range: \(x=-1,x\leq \frac{7}{2}\) - step8: Find the intersection: \(x=-1\) - step9: Check the solution: \(x=-1\) Solve the equation \( 2 x^{2}-3 x-4=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(2x^{2}-3x-4=0\) - step1: Solve using the quadratic formula: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 2\left(-4\right)}}{2\times 2}\) - step2: Simplify the expression: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 2\left(-4\right)}}{4}\) - step3: Simplify the expression: \(x=\frac{3\pm \sqrt{41}}{4}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{3+\sqrt{41}}{4}\\&x=\frac{3-\sqrt{41}}{4}\end{align}\) - step5: Rewrite: \(x_{1}=\frac{3-\sqrt{41}}{4},x_{2}=\frac{3+\sqrt{41}}{4}\) Solve the equation \( x(x-4) - 5 \geq 0 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(x\left(x-4\right)-5\geq 0\) - step1: Rearrange the terms: \(x^{2}-4x-5\geq 0\) - step2: Rewrite the expression: \(x^{2}-4x-5=0\) - step3: Factor the expression: \(\left(x-5\right)\left(x+1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&x-5=0\\&x+1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=5\\&x=-1\end{align}\) - step6: Determine the test intervals: \(\begin{align}&x<-1\\&-15\end{align}\) - step7: Choose a value: \(\begin{align}&x_{1}=-2\\&x_{2}=2\\&x_{3}=6\end{align}\) - step8: Test the chosen value: \(\begin{align}&x<-1\textrm{ }\textrm{is the solution}\\&-15\textrm{ }\textrm{is the solution}\end{align}\) - step9: Include the critical value: \(\begin{align}&x\leq -1\textrm{ }\textrm{is the solution}\\&x\geq 5\textrm{ }\textrm{is the solution}\end{align}\) - step10: The final solution is \(x \in \left(-\infty,-1\right]\cup \left[5,+\infty\right):\) \(x \in \left(-\infty,-1\right]\cup \left[5,+\infty\right)\) Let's solve each equation step by step. ### 1.1. \( 111 x^{2}+5 x-14=0 \) Using the quadratic formula, we find the roots of the equation: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 111 \), \( b = 5 \), and \( c = -14 \). Calculating the discriminant: \[ b^2 - 4ac = 5^2 - 4 \cdot 111 \cdot (-14) = 25 + 6224 = 6249 \] Now, substituting into the quadratic formula: \[ x = \frac{-5 \pm \sqrt{6249}}{2 \cdot 111} \] Calculating the roots gives us: \[ x_{1} = -\frac{14}{37}, \quad x_{2} = \frac{1}{3} \] ### 1.2. \( 2 x^{2}-3 x-4=0 \) Using the quadratic formula again: Here, \( a = 2 \), \( b = -3 \), and \( c = -4 \). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-4) = 9 + 32 = 41 \] Now, substituting into the quadratic formula: \[ x = \frac{3 \pm \sqrt{41}}{4} \] Calculating the roots gives us: \[ x_{1} \approx -0.85, \quad x_{2} \approx 2.35 \] ### 1.3. \( x(x-4) \geq 5 \) Rearranging the inequality: \[ x^2 - 4x - 5 \geq 0 \] Factoring gives: \[ (x - 5)(x + 1) \geq 0 \] The critical points are \( x = -1 \) and \( x = 5 \). Testing intervals: - For \( x < -1 \): both factors are negative, product is positive. - For \( -1 < x < 5 \): one factor is negative, product is negative. - For \( x > 5 \): both factors are positive, product is positive. Thus, the solution is: \[ x \in (-\infty, -1] \cup [5, +\infty) \] ### 1.4. \( \sqrt{7-2 x}-x-4=0 \) Rearranging gives: \[ \sqrt{7-2x} = x + 4 \] Squaring both sides: \[ 7 - 2x = (x + 4)^2 \] Expanding and rearranging: \[ 7 - 2x = x^2 + 8x + 16 \] \[ 0 = x^2 + 10x + 9 \] Factoring gives: \[ (x + 1)(x + 9) = 0 \] Thus, the solutions are: \[ x = -1 \quad \text{(valid since it satisfies the original equation)} \] ### Summary of Solutions: 1. \( 111 x^{2}+5 x-14=0 \) gives \( x_{1} = -\frac{14}{37}, x_{2} = \frac{1}{3} \) 2. \( 2 x^{2}-3 x-4=0 \) gives \( x_{1} \approx -0.85, x_{2} \approx 2.35 \) 3. \( x(x-4) \geq 5 \) gives \( x \in (-\infty, -1] \cup [5, +\infty) \) 4. \( \sqrt{7-2 x}-x-4=0 \) gives \( x = -1 \)