1.1 Solve for \( x \) in the following equations. \( \begin{array}{l}1.1 .1 \frac{x-1}{2}-\frac{3 x}{4}<-6 \\ 1.1 .2(1-4 x)(-\sqrt{x}+5)=0 \\ 1.2 \text { Solve for } x \text { by completing the square: } \\ a x^{2}+b x+c=0 \\ 1.3 \text { Solve for } x \text { by using quadratic formula: } \\ x^{2}+9 x-36=0\end{array} \)

**1.1.1** We start with the inequality \[ \frac{x-1}{2}-\frac{3x}{4}<-6. \] Write both terms with the common denominator \(4\): \[ \frac{2(x-1)-3x}{4}=\frac{2x-2-3x}{4}=\frac{-x-2}{4}. \] So the inequality becomes \[ \frac{-x-2}{4} < -6. \] Multiply both sides by \(4\) (note that \(4>0\) so the inequality sign remains unchanged): \[ -x-2 < -24. \] Add \(2\) to both sides: \[ -x < -22. \] Multiply by \(-1\) (remember to reverse the inequality sign): \[ x > 22. \] --- **1.1.2** The equation is \[ (1-4x)(-\sqrt{x}+5)=0. \] Using the zero product property, set each factor equal to zero. *First factor:* \[ 1-4x=0 \quad \Rightarrow \quad 4x=1 \quad \Rightarrow \quad x=\frac{1}{4}. \] *Second factor:* \[ -\sqrt{x}+5=0 \quad \Rightarrow \quad \sqrt{x}=5 \quad \Rightarrow \quad x=25. \] Thus, the solutions are \[ x=\frac{1}{4} \quad \text{or} \quad x=25. \] --- **1.2** To solve the general quadratic \[ ax^{2}+bx+c=0 \] by completing the square, follow these steps: 1. Divide the entire equation by \(a\) (assuming \(a\ne0\)): \[ x^{2}+\frac{b}{a}x+\frac{c}{a}=0 \quad \Longrightarrow \quad x^{2}+\frac{b}{a}x=-\frac{c}{a}. \] 2. Add \(\left(\frac{b}{2a}\right)^{2}=\frac{b^2}{4a^2}\) to both sides to complete the square: \[ x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}=-\frac{c}{a}+\frac{b^{2}}{4a^{2}}. \] 3. Write the left side as a perfect square: \[ \left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}. \] 4. Take the square root of both sides: \[ x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}. \] 5. Solve for \(x\): \[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \] --- **1.3** For the quadratic \[ x^{2}+9x-36=0, \] identify \(a=1\), \(b=9\), and \(c=-36\). Use the quadratic formula: \[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \] Substitute the values: \[ x=\frac{-9\pm\sqrt{9^{2}-4\cdot1\cdot(-36)}}{2\cdot1}=\frac{-9\pm\sqrt{81+144}}{2}=\frac{-9\pm\sqrt{225}}{2}. \] Since \(\sqrt{225}=15\), we get two solutions: 1. \(x=\frac{-9+15}{2}=\frac{6}{2}=3,\) 2. \(x=\frac{-9-15}{2}=\frac{-24}{2}=-12.\) Thus, the solutions are \[ x=3 \quad \text{or} \quad x=-12. \]