1. **Solve $8x^{2} + 6x - 2 = 0$ using quadratic formula.** The quadratic formula is given by: $$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$$ Here, $a=8$, $b=6$, $c=-2$. Calculate the discriminant: $$\Delta = b^{2} - 4ac = 6^{2} - 4 \times 8 \times (-2) = 36 + 64 = 100$$ Find the roots: $$x = \frac{-6 \pm \sqrt{100}}{2 \times 8} = \frac{-6 \pm 10}{16}$$ So, $$x_1 = \frac{-6 + 10}{16} = \frac{4}{16} = \frac{1}{4}$$ $$x_2 = \frac{-6 - 10}{16} = \frac{-16}{16} = -1$$ 2. **Solve the equation $\sqrt{8x} + 5 = 4x - 5$.** Rewrite as: $$\sqrt{8x} = 4x - 10$$ Note the right side must be $2geq 0$
. Square both sides: $$8x = (4x - 10)^{2} = 16x^{2} - 80x + 100$$ Bring all terms to one side: $$16x^{2} - 80x + 100 - 8x = 0 \Rightarrow 16x^{2} - 88x + 100 = 0$$ Divide entire equation by 4: $$4x^{2} - 22x + 25 = 0$$ Use quadratic formula: $a=4$, $b=-22$, $c=25$ Calculate discriminant: $$\Delta = (-22)^{2} - 4 \times 4 \times 25 = 484 - 400 = 84$$ Roots are: $$x = \frac{22 \pm \sqrt{84}}{8} = \frac{22 \pm 2\sqrt{21}}{8} = \frac{11 \pm \sqrt{21}}{4}$$ Approximate values: $x_1 = \frac{11 + 4.58}{4} = 3.895$, $x_2 = \frac{11 - 4.58}{4} = 1.605$ Check in original equation to avoid extraneous roots: For $3.895$, LHS $= \sqrt{8 \times 3.895} + 5 \approx \sqrt{31.16} + 5 = 5.58 + 5 = 10.58$; RHS $= 4 \times 3.895 - 5 = 15.58 - 5 = 10.58$ valid. For $1.605$, LHS $= \sqrt{12.84} + 5 = 3.58 + 5 = 8.58$; RHS $= 4 \times 1.605 - 5 = 6.42 - 5 = 1.42$ invalid. So, $x = \frac{11 + \sqrt{21}}{4}$ 3. **Examine the nature of roots of $4x^2 + 2x - 2 = 0$ and solve it.** Calculate discriminant: $$\Delta = 2^2 - 4 \times 4 \times (-2) = 4 + 32 = 36$$ Since $$\Delta > 0$$, roots are real and distinct. Roots: $$x = \frac{-2 \pm 6}{2 \times 4} = \frac{-2 \pm 6}{8}$$ $$x_1 = \frac{-2 + 6}{8} = \frac{4}{8} = \frac{1}{2}$$ $$x_2 = \frac{-2 - 6}{8} = \frac{-8}{8} = -1$$ 4. **Find $k$ such that product of roots of $2x^{2} - kx + 4 = 0$ equals twice the sum of roots.** Sum of roots: $$\alpha + \beta = \frac{k}{2}$$ Product of roots: $$\alpha \beta = \frac{4}{2} = 2$$ Given: $$2 = 2 \times \frac{k}{2} \Rightarrow 2 = k$$ So, $k = 2$ 5. **Prove if $\frac{2a - 5b}{2a + 5b} = \frac{2c - 5d}{2c + 5d}$ then $\frac{a}{b} = \frac{c}{d}$.** Cross multiply: $$(2a - 5b)(2c + 5d) = (2c - 5d)(2a + 5b)$$ Expand both sides: LHS: $$4ac + 10ad - 10bc - 25bd$$ RHS: $$4ac + 10bc - 10ad - 25bd$$ Bring terms on one side: $$4ac + 10ad - 10bc - 25bd - 4ac - 10bc + 10ad + 25bd = 0$$ Simplify: $$10ad + 10ad - 10bc - 10bc - 25bd + 25bd = 0 \\ 20ad - 20bc = 0$$ Divide by 20: $$ad = bc$$ Divide both sides by $bd$ (assuming $b,d \neq 0$): $$\frac{a}{b} = \frac{c}{d}$$ 6. **Resolve partial fraction: $\frac{3x+1}{(x+1)(2x-3)}$.** Write: $$\frac{3x+1}{(x+1)(2x-3)} = \frac{A}{x+1} + \frac{B}{2x-3}$$ Multiply both sides by denominator: $$3x + 1 = A(2x - 3) + B(x + 1)$$ Expand: $$3x + 1 = 2Ax - 3A + Bx + B$$ Collect like terms: $$3x + 1 = (2A + B)x + (-3A + B)$$ Equate coefficients: From $x$: $$3 = 2A + B$$ Constant: $$1 = -3A + B$$ Solve system: From first: $$B = 3 - 2A$$ Substitute into second: $$1 = -3A + 3 - 2A = 3 - 5A \Rightarrow 5A = 2 \Rightarrow A = \frac{2}{5}$$ Then: $$B = 3 - 2 \times \frac{2}{5} = 3 - \frac{4}{5} = \frac{15}{5} - \frac{4}{5} = \frac{11}{5}$$ So: $$\frac{3x+1}{(x+1)(2x-3)} = \frac{2/5}{x+1} + \frac{11/5}{2x-3}$$ 6b. We may write: $$= \frac{2}{5(x+1)} + \frac{11}{5(2x-3)}$$ 7. **Prove: $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ where $A=\{a,b,c\}$, $B=\{b,c,d\}$, $C=\{c,d,e\}$.** Compute $B \cup C = \{b,c,d,e\}$ Then $A \cap (B \cup C) = \{b,c\}$ Compute: $$A \cap B = \{b,c\}, \quad A \cap C = \{c\}$$ Their union: $$(A \cap B) \cup (A \cap C) = \{b,c\} \cup \{c\} = \{b,c\}$$ Hence shown: $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$ 8. **Find Harmonic Mean for Classes with frequencies:** Classes: $0-15, 15-30, 30-45, 45-60$ Frequencies: $3, 8, 6, 5$ Find class midpoints: $$7.5, 22.5, 37.5, 52.5$$ Formula for Harmonic Mean (HM): $$HM = \frac{\sum f_i}{\sum \frac{f_i}{x_i}}$$ Calculate denominator: $$\sum \frac{f_i}{x_i} = \frac{3}{7.5} + \frac{8}{22.5} + \frac{6}{37.5} + \frac{5}{52.5}$$ Calculations: $$\frac{3}{7.5} = 0.4, \quad \frac{8}{22.5} \approx 0.3556, \quad \frac{6}{37.5} = 0.16, \quad \frac{5}{52.5} \approx 0.0952$$ Sum: $$0.4 + 0.3556 + 0.16 + 0.0952 = 1.0108$$ Sum of frequencies: $$3 + 8 + 6 + 5 = 22$$ Thus, $$HM = \frac{22}{1.0108} \approx 21.77$$ 9. **If $\sin \theta = -\frac{3}{5}$ and $\theta$ lies in the 4th quadrant, find other ratios.** In 4th quadrant: $sin<0$ and $cos>0$ Use Pythagoras: $$\cos \theta = +\sqrt{1 - \sin^2 \theta} = +\sqrt{1 - \left(\frac{-3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$ Calculate: $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-3/5}{4/5} = -\frac{3}{4}$$ $$\csc \theta = \frac{1}{\sin \theta} = -\frac{5}{3}$$ $$\sec \theta = \frac{1}{\cos \theta} = \frac{5}{4}$$ $$\cot \theta = \frac{1}{\tan \theta} = -\frac{4}{3}$$ 10. **Prove $\sin^2 \theta - \cos^2 \theta = 1 - 2 \cos^2 \theta$.** Start with LHS: $$\sin^2 \theta - \cos^2 \theta$$ Use identity: $$\sin^2 \theta = 1 - \cos^2 \theta$$ So, $$1 - \cos^2 \theta - \cos^2 \theta = 1 - 2 \cos^2 \theta$$ Hence proved. 11. **If lengths of two tangents to a circle from external point are $x^2 + 6$ and $5x$, find $x$.** Since lengths of tangents from the same point to a circle are equal: $$x^2 + 6 = 5x$$ Arrange: $$x^2 - 5x + 6 = 0$$ Factor: $$(x - 2)(x - 3) = 0$$ Solutions: $$x = 2, \quad x=3$$ 12. **Prove equal chords of congruent circles have congruent corresponding arcs.** Equal chords in congruent circles subtend equal angles at the centers. Since both circles are congruent, their radii are equal. Equal chords subtend equal central angles. Equal central angles mean arcs subtended by these chords are equal (congruent). Hence, minor, major or semicircular arcs corresponding to equal chords are congruent. **Section-C** 13. **Using Apollonius' theorem, find lengths of medians of triangle with sides 8 cm, 15 cm, 17 cm.** Apollonius' theorem: $$m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$ Label sides: $$a=8, b=15, c=17$$ Calculate medians: $$m_a^2 = \frac{2 \times 15^2 + 2 \times 17^2 - 8^2}{4} = \frac{2 \times 225 + 2 \times 289 - 64}{4} = \frac{450 + 578 - 64}{4} = \frac{964}{4} = 241$$ $$m_a = \sqrt{241} \approx 15.52$$ Similarly, for $m_b$: $$m_b^2 = \frac{2a^2 + 2c^2 - b^2}{4} = \frac{2 \times 8^2 + 2 \times 17^2 - 15^2}{4} = \frac{128 + 578 - 225}{4} = \frac{481}{4} = 120.25$$ $$m_b = \sqrt{120.25} = 10.97$$ For $m_c$: $$m_c^2 = \frac{2a^2 + 2b^2 - c^2}{4} = \frac{2 \times 8^2 + 2 \times 15^2 - 17^2}{4} = \frac{128 + 450 - 289}{4} = \frac{289}{4} = 72.25$$ $$m_c = \sqrt{72.25} = 8.5$$ 14. **Prove only one circle passes through three non-collinear points.** Three non-collinear points define a unique triangle. Perpendicular bisectors of any two sides intersect at one point, the circumcenter. This point is equidistant from all three vertices. This defines a unique circle passing through all three points. Hence, only one such circle exists. 15. **Prove that any two angles in the same segment of a circle are equal.** Angles subtended by the same chord in the same segment are equal. These angles subtend equal arcs and share the same chord as base. Hence, by circle geometry, these angles are equal.