1. **Solve 6x² + 5x + 1 = 0 using quadratic formula** The quadratic formula is given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=6$, $b=5$, and $c=1$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 5^2 - 4 \cdot 6 \cdot 1 = 25 - 24 = 1$$ Since $\Delta > 0$, two distinct real roots exist. Calculate the roots: $$x = \frac{-5 \pm \sqrt{1}}{12} = \frac{-5 \pm 1}{12}$$ So, - $x_1 = \frac{-5 + 1}{12} = \frac{-4}{12} = -\frac{1}{3}$ - $x_2 = \frac{-5 - 1}{12} = \frac{-6}{12} = -\frac{1}{2}$ 2. **Solve the equation $\sqrt{3}x + 5 + 2 = 4x$** First simplify the equation: $$\sqrt{3} x + 7 = 4x$$ Bring terms with $x$ to one side: $$4x - \sqrt{3} x = 7$$ Factor out $x$: $$x(4 - \sqrt{3}) = 7$$ Solve for $x$: $$x = \frac{7}{4 - \sqrt{3}}$$ Rationalize the denominator: $$x = \frac{7}{4 - \sqrt{3}} \times \frac{4 + \sqrt{3}}{4 + \sqrt{3}} = \frac{7(4 + \sqrt{3})}{(4)^2 - (\sqrt{3})^2} = \frac{7(4 + \sqrt{3})}{16 - 3} = \frac{7(4 + \sqrt{3})}{13}$$ 3. **Determine the nature and solve 7x² + 2x - 1 = 0** Compute discriminant: $$\Delta = 2^2 - 4 \cdot 7 \cdot (-1) = 4 + 28 = 32 > 0$$ So two distinct real roots exist. Roots: $$x = \frac{-2 \pm \sqrt{32}}{14} = \frac{-2 \pm 4\sqrt{2}}{14} = \frac{-1 \pm 2\sqrt{2}}{7}$$ 4. **Find $k$ such that product of roots = 2 × sum of roots for $2x^2 - kx + 4 = 0$** Sum of roots: $$S = \frac{k}{2}$$ Product of roots: $$P = \frac{4}{2} = 2$$ Given: $$P = 2S \Rightarrow 2 = 2 \times \frac{k}{2} = k$$ So, $k = 2$. 5. **Prove if $\frac{3a - 7b}{3a + 7b} = \frac{3c - 7d}{3c + 7d}$, then $\frac{a}{b} = \frac{c}{d}$** Cross multiply: $$(3a - 7b)(3c + 7d) = (3c - 7d)(3a + 7b)$$ Expand both sides: $$9ac + 21ad - 21bc - 49bd = 9ac + 21bc - 21ad - 49bd$$ Cancel common terms: $$21ad - 21bc = 21bc - 21ad$$ Bring terms to one side: $$21ad - 21bc - 21bc + 21ad = 0 \Rightarrow 42ad - 42bc = 0$$ Divide both sides by 42: $$ad - bc = 0 \Rightarrow ad = bc$$ Divide both sides by $bd$ (assuming $b, d \neq 0$): $$\frac{a}{b} = \frac{c}{d}$$ 6. **Partial fraction decomposition of $ \frac{3x+1}{(x+1)(2x^2 - 3)} $** Assume: $$\frac{3x+1}{(x+1)(2x^2 -3)} = \frac{A}{x+1} + \frac{Bx + C}{2x^2 - 3}$$ Multiply both sides by the denominator: $$3x + 1 = A(2x^2 -3) + (Bx + C)(x + 1)$$ Expand: $$3x + 1 = 2Ax^2 - 3A + Bx^2 + Bx + Cx + C$$ Group terms: $$3x + 1 = (2A + B) x^2 + (B + C) x + (-3A + C)$$ Equate coefficients: - Coefficient of $x^2$: $0 = 2A + B$ - Coefficient of $x$: $3 = B + C$ - Constant term: $1 = -3A + C$ From first: $B = -2A$ From second: $3 = -2A + C => C = 3 + 2A$ From third: $1 = -3A + C = -3A + 3 + 2A = 3 - A$ So, $1 = 3 - A \Rightarrow A = 2$ Then $B = -4$ and $C = 3 + 4 = 7$ Final decomposition: $$\frac{3x+1}{(x+1)(2x^2 - 3)} = \frac{2}{x+1} + \frac{-4x + 7}{2x^2 - 3}$$ 7. **Prove $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ for** $A=\{a,b,c\}$, $B=\{b,c,d\}$, $C=\{c,d,e\}$ - $B \cup C = \{b,c,d,e\}$ - $A \cap (B \cup C) = \{b,c\}$ - $A \cap B = \{b,c\}$ - $A \cap C = \{c\}$ - $(A \cap B) \cup (A \cap C) = \{b,c\} \cup \{c\} = \{b,c\}$ Both sides equal $\{b,c\}$ thus proven. 8. **Find harmonic mean of frequencies: 3,8,6,5** Harmonic mean (HM): $$HM = \frac{n}{\sum \frac{1}{x_i}}$$ Where $n=4$, values: $3,8,6,5$ Calculate sum of reciprocals: $$\frac{1}{3} + \frac{1}{8} + \frac{1}{6} + \frac{1}{5} = 0.3333 + 0.125 + 0.1667 + 0.2 = 0.825$$ Then $$HM = \frac{4}{0.825} \approx 4.85$$ 9. **If $\cos \theta = -\frac{3}{5}$ in 2nd quadrant, find other ratios** In 2nd quadrant, $\sin \theta > 0$, $\cos \theta < 0$. Calculate $\sin \theta$ using: $$\sin^2 \theta + \cos^2 \theta = 1$$ So, $$\sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\sin \theta = +\frac{4}{5}$$ Find $\tan \theta$: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{4/5}{-3/5} = -\frac{4}{3}$$ Cosecant: $$\csc \theta = \frac{1}{\sin \theta} = \frac{5}{4}$$ Secant: $$\sec \theta = \frac{1}{\cos \theta} = -\frac{5}{3}$$ Cotangent: $$\cot \theta = \frac{1}{\tan \theta} = -\frac{3}{4}$$ 10. **Prove $ (1 + \sin \theta)(1 - \sin \theta) = \frac{1}{\sec^2 \theta} $** Left Side: $$(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2 \theta = \cos^2 \theta$$ Right Side: $$\frac{1}{\sec^2 \theta} = \cos^2 \theta$$ Left Side = Right Side, hence proved. 11. **If lengths of two tangents from external point are $x^2 -5$ and 21, find $x$** Length of tangents from one point are equal, so $$x^2 -5 = 21$$ $$x^2 = 26$$ $$x = \pm \sqrt{26}$$ 12. **Prove that equal chords of congruent circles subtend equal angles at centers** - Congruent circles have same radius. - Equal chords in circles with same radius subtend equal arcs. - Equal arcs subtend equal angles at their centers. Thus, equal chords of congruent circles subtend equal angles at centers. 13. **Find medians of triangle with sides 8cm, 15cm, 17cm using Apollonius' theorem** Apollonius' theorem states: $$m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$ Label triangle with sides $a=8$, $b=15$, $c=17$. Median from $a$: $$m_a^2 = \frac{2 \times 15^2 + 2 \times 17^2 - 8^2}{4} = \frac{2 \times 225 + 2 \times 289 - 64}{4} = \frac{450 + 578 - 64}{4} = \frac{964}{4} = 241$$ $$m_a = \sqrt{241} \approx 15.52 \text{ cm}$$ Median from $b$: $$m_b^2 = \frac{2 \times 8^2 + 2 \times 17^2 - 15^2}{4} = \frac{2 \times 64 + 2 \times 289 - 225}{4} = \frac{128 + 578 - 225}{4} = \frac{481}{4} = 120.25$$ $$m_b = \sqrt{120.25} = 10.97 \text{ cm}$$ Median from $c$: $$m_c^2 = \frac{2 \times 8^2 + 2 \times 15^2 - 17^2}{4} = \frac{2 \times 64 + 2 \times 225 - 289}{4} = \frac{128 + 450 - 289}{4} = \frac{289}{4} = 72.25$$ $$m_c = \sqrt{72.25} = 8.5 \text{ cm}$$ 14. **Prove if two chords of a circle are congruent, then they are equidistant from the center** - Draw radii to midpoints of chords; these radii are perpendicular bisectors. - Equal chords have equal lengths. - In the triangle formed by the radius, chord, and line from center to chord midpoint, the perpendicular distance from center to chord is same for equal chords. Therefore, equal chords are equidistant from the center. 15. **Prove opposite angles of quadrilateral inscribed in a circle are supplementary** - Let quadrilateral $ABCD$ be inscribed in a circle. - Angle $A$ and angle $C$ subtend arcs $BCD$ and $DAB$ respectively. - Arcs $BCD$ and $DAB$ together cover the circle, so sum of arcs is 360°. - Measure of angle equals half measure of intercepted arc. Thus, $$\angle A + \angle C = \frac{1}{2}(\text{arc } BCD + \text{arc } DAB) = \frac{1}{2} \times 360° = 180°$$ Similarly for angles $B$ and $D$. 16. **Inscribe a square in a circle of radius 8cm** - Diagonal of square equals diameter of circle: $$d = 2r = 16 cm$$ - Side length $s$ of square satisfies: $$s \sqrt{2} = d = 16$$ - Solve for $s$: $$s = \frac{16}{\sqrt{2}} = 8\sqrt{2} \approx 11.31 cm$$