Question 1(a) - Solving Equations

i) $\frac{x^2 - 169}{x^2} = 0$

First, we can simplify the equation by eliminating the denominator: $\frac{x^2 - 169}{x^2} = 0 \implies x^2 - 169 = 0$ Next, solve for $x$: $x^2 - 169 = 0 \implies x^2 = 169 \implies x = \pm \sqrt{169} \implies x = \pm 13$

Solution: $x = 13$ or $x = -13$

ii) $2x^2 - 7x + 6 = 0$ (Using the quadratic formula)

The quadratic formula is given by: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For the equation $2x^2 - 7x + 6 = 0$, we have $a = 2$, $b = -7$, and $c = 6$. Plugging these values into the quadratic formula: $x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 - 48}}{4} = \frac{7 \pm 1}{4}$ This gives us two solutions: $x = \frac{7 + 1}{4} = 2 \quad \text{and} \quad x = \frac{7 - 1}{4} = \frac{6}{4} = \frac{3}{2}$

Solution: $x = 2$ or $x = \frac{3}{2}$

iii) $x^2 - 12x + 5 = 7$ (Applying the square root property)

First, simplify the equation: $x^2 - 12x + 5 = 7 \implies x^2 - 12x + 5 - 7 = 0 \implies x^2 - 12x - 2 = 0$ This is a quadratic equation, and it can be solved using the quadratic formula: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{12 \pm \sqrt{144 + 8}}{2} = \frac{12 \pm \sqrt{152}}{2} = \frac{12 \pm 2\sqrt{38}}{2} = 6 \pm \sqrt{38}$

Solution: $x = 6 + \sqrt{38}$ or $x = 6 - \sqrt{38}$

Question 1(b) - Solving Inequalities

ii) (5 \left|\frac{2x+3