1. **Problem 1.1.1: Solve the quadratic equation** $$x^2 - 4x + 3 = 0$$ 2. **Formula:** For quadratic equations $$ax^2 + bx + c = 0$$, solutions are given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. **Apply values:** Here, $a=1$, $b=-4$, $c=3$. 4. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-4)^2 - 4 \times 1 \times 3 = 16 - 12 = 4$$ 5. Calculate roots: $$x = \frac{-(-4) \pm \sqrt{4}}{2 \times 1} = \frac{4 \pm 2}{2}$$ 6. Roots: $$x_1 = \frac{4 + 2}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{4 - 2}{2} = \frac{2}{2} = 1$$ --- 1. **Problem 1.1.2: Solve the quadratic equation** $$5x^2 - 5x + 1 = 0$$ to two decimal places. 2. **Apply formula:** $a=5$, $b=-5$, $c=1$. 3. Calculate discriminant: $$\Delta = (-5)^2 - 4 \times 5 \times 1 = 25 - 20 = 5$$ 4. Calculate roots: $$x = \frac{-(-5) \pm \sqrt{5}}{2 \times 5} = \frac{5 \pm \sqrt{5}}{10}$$ 5. Approximate roots: $$\sqrt{5} \approx 2.236$$ $$x_1 = \frac{5 + 2.236}{10} = \frac{7.236}{10} = 0.72$$ $$x_2 = \frac{5 - 2.236}{10} = \frac{2.764}{10} = 0.28$$ --- 1. **Problem 1.1.3: Solve the inequality** $$x^2 - 3x - 10 > 0$$ 2. Factor the quadratic: $$x^2 - 3x - 10 = (x - 5)(x + 2)$$ 3. The inequality is: $$(x - 5)(x + 2) > 0$$ 4. The product is positive when both factors are positive or both are negative. 5. Case 1: Both positive $$x - 5 > 0 \Rightarrow x > 5$$ $$x + 2 > 0 \Rightarrow x > -2$$ Since $x > 5$ implies $x > -2$, solution here is: $$x > 5$$ 6. Case 2: Both negative $$x - 5 < 0 \Rightarrow x < 5$$ $$x + 2 < 0 \Rightarrow x < -2$$ Since $x < -2$ implies $x < 5$, solution here is: $$x < -2$$ 7. **Final solution:** $$x < -2 \quad \text{or} \quad x > 5$$ --- 1. **Problem 1.1.4: Solve the equation** $$3\sqrt{x} = x - 4$$ 2. Let $y = \sqrt{x}$, so $x = y^2$. 3. Substitute: $$3y = y^2 - 4$$ 4. Rearrange: $$y^2 - 3y - 4 = 0$$ 5. Solve quadratic for $y$: $$a=1, b=-3, c=-4$$ 6. Discriminant: $$\Delta = (-3)^2 - 4 \times 1 \times (-4) = 9 + 16 = 25$$ 7. Roots: $$y = \frac{3 \pm 5}{2}$$ 8. Calculate roots: $$y_1 = \frac{3 + 5}{2} = \frac{8}{2} = 4$$ $$y_2 = \frac{3 - 5}{2} = \frac{-2}{2} = -1$$ 9. Since $y = \sqrt{x} \geq 0$, discard $y = -1$. 10. So $y = 4$, then: $$x = y^2 = 4^2 = 16$$ 11. **Check solution:** $$3\sqrt{16} = 3 \times 4 = 12$$ $$16 - 4 = 12$$ Both sides equal, so $x=16$ is valid. **Final answers:** 1.1.1: $x=3$ or $x=1$ 1.1.2: $x=0.72$ or $x=0.28$ 1.1.3: $x < -2$ or $x > 5$ 1.1.4: $x=16$