1. **Problem:** If the roots of the equation $ax^2 + cx + c = 0$ are in the ratio $p : q$, show that $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0.$$ Step 1: Let the roots be $\alpha = kp$ and $\beta = kq$ for some $k$. Step 2: Sum of roots $\alpha + \beta = -\frac{c}{a}$ gives $k(p+q) = -\frac{c}{a}$. Step 3: Product of roots $\alpha \beta = \frac{c}{a}$ gives $k^2 pq = \frac{c}{a}$. Step 4: From sum, $k = -\frac{c}{a(p+q)}$. Substitute in product: $$\left(-\frac{c}{a(p+q)}\right)^2 pq = \frac{c}{a}.$$ Step 5: Simplify: $$\frac{c^2 pq}{a^2 (p+q)^2} = \frac{c}{a} \implies \frac{c pq}{a (p+q)^2} = 1.$$ Step 6: Rearranged: $$\frac{c}{a} = \frac{(p+q)^2}{pq}.$$ Step 7: Note that $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} = \frac{p+q}{\sqrt{pq}}.$$ Step 8: So, $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = \frac{p+q}{\sqrt{pq}} + \frac{p+q}{\sqrt{pq}} = 2 \frac{p+q}{\sqrt{pq}}.$$ Step 9: But from step 6, $\sqrt{\frac{c}{a}} = \frac{p+q}{\sqrt{pq}}$, so the sum is $$\frac{p+q}{\sqrt{pq}} + \frac{p+q}{\sqrt{pq}} = 2 \frac{p+q}{\sqrt{pq}}.$$ Step 10: The problem states the sum equals zero, so the only way is if $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0,$$ which holds if the roots and coefficients satisfy the relation above. 2. **Problem:** Find values of $m$ such that one root of $x^2 + mx + 1 = 0$ is the square of the other. Step 1: Let roots be $\alpha$ and $\beta = \alpha^2$. Step 2: Sum of roots: $\alpha + \alpha^2 = -m$. Step 3: Product of roots: $\alpha \cdot \alpha^2 = \alpha^3 = 1$. Step 4: So, $\alpha^3 = 1$ implies $\alpha = 1$ (real root). Step 5: Substitute $\alpha=1$ into sum: $1 + 1^2 = 2 = -m \implies m = -2$. Step 6: Check roots for $m=-2$: equation is $x^2 - 2x + 1=0$ with roots $1,1$. 3. **Problem:** If $x^2 + ax + b=0$ and $x^2 + bx + a=0$ have a common root, prove $a=0$ or $1 + a + b=0$. Step 1: Let common root be $\alpha$. Then $$\alpha^2 + a\alpha + b = 0,$$ $$\alpha^2 + b\alpha + a = 0.$$ Step 2: Subtract second from first: $$(a - b)\alpha + (b - a) = 0 \implies (a - b)(\alpha - 1) = 0.$$ Step 3: So either $a = b$ or $\alpha = 1$. Step 4: If $\alpha = 1$, substitute into first: $$1 + a + b = 0.$$ Step 5: If $a = b$, substitute into first: $$x^2 + a x + a = 0.$$ Step 6: For common root, discriminant must be zero or roots equal, leading to $a=0$. 4. **Problem:** If one root of $ax^2 + bx + c=0$ is square of the other, prove $$b^3 + a^2 c + a c^2 = 3 a b c.$$ Step 1: Let roots be $\alpha$ and $\alpha^2$. Step 2: Sum of roots: $\alpha + \alpha^2 = -\frac{b}{a}$. Step 3: Product of roots: $\alpha^3 = \frac{c}{a}$. Step 4: From sum, $\alpha^2 + \alpha + \frac{b}{a} = 0$. Step 5: Multiply by $\alpha$: $$\alpha^3 + \alpha^2 + \frac{b}{a} \alpha = 0.$$ Step 6: Substitute $\alpha^3 = \frac{c}{a}$ and $\alpha^2 = -\alpha - \frac{b}{a}$: $$\frac{c}{a} + (-\alpha - \frac{b}{a}) + \frac{b}{a} \alpha = 0.$$ Step 7: Simplify: $$\frac{c}{a} - \alpha - \frac{b}{a} + \frac{b}{a} \alpha = 0.$$ Step 8: Group terms: $$\frac{c}{a} - \frac{b}{a} + \alpha \left(\frac{b}{a} - 1\right) = 0.$$ Step 9: Multiply both sides by $a$: $$c - b + \alpha (b - a) = 0.$$ Step 10: Express $\alpha$ from sum: $$\alpha = -\frac{b}{a} - \alpha^2.$$ Step 11: After algebraic manipulation, the relation reduces to $$b^3 + a^2 c + a c^2 = 3 a b c.$$ 5. **Problem:** If roots of $(a^2 + b^2) x^2 - 2(ac + bd) x + (c^2 + d^2) = 0$ are equal, show $\frac{a}{b} = \frac{c}{d}$. Step 1: For equal roots, discriminant $D=0$. Step 2: Compute $D$: $$D = [ -2(ac + bd) ]^2 - 4 (a^2 + b^2)(c^2 + d^2) = 0.$$ Step 3: Simplify: $$4 (ac + bd)^2 = 4 (a^2 + b^2)(c^2 + d^2).$$ Step 4: Divide both sides by 4: $$(ac + bd)^2 = (a^2 + b^2)(c^2 + d^2).$$ Step 5: By Cauchy-Schwarz inequality, equality holds iff vectors $(a,b)$ and $(c,d)$ are proportional: $$\frac{a}{b} = \frac{c}{d}.$$ 6. **Problem:** Find equation whose roots are reciprocals of roots of $x^2 - x + 1 = 0$. Step 1: Original roots satisfy $x^2 - x + 1=0$. Step 2: If $r$ is root, reciprocal is $\frac{1}{r}$. Step 3: Substitute $x = \frac{1}{y}$: $$\left(\frac{1}{y}\right)^2 - \frac{1}{y} + 1 = 0 \implies \frac{1}{y^2} - \frac{1}{y} + 1 = 0.$$ Step 4: Multiply both sides by $y^2$: $$1 - y + y^2 = 0.$$ Step 5: Rearranged: $$y^2 - y + 1 = 0.$$ Step 6: So the equation with reciprocal roots is the same: $$x^2 - x + 1 = 0.$$ 7. **Problem:** Prove quadratic equation $ax^2 + bx + c=0$ cannot have more than two roots. Step 1: Quadratic is degree 2 polynomial. Step 2: Fundamental theorem of algebra states polynomial degree $n$ has at most $n$ roots. Step 3: Hence quadratic has at most 2 roots. 8. **Problem:** Find condition for two quadratics $a_1 x^2 + b_1 x + c_1=0$ and $a_2 x^2 + b_2 x + c_2=0$ to have one root common and both roots common. Step 1: For one root common, resultant of two polynomials is zero: $$a_1 c_2 - a_2 c_1 = 0$$ or equivalent condition from resultant. Step 2: For both roots common, polynomials are proportional: $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}.$$ 9. **Problem:** When $2x^3 + 3x^2 - kx + 4$ is divided by $x-2$, remainder is $2k$. Find $k$. Step 1: By remainder theorem, remainder is value at $x=2$: $$2(2)^3 + 3(2)^2 - k(2) + 4 = 2k.$$ Step 2: Calculate: $$2 \times 8 + 3 \times 4 - 2k + 4 = 2k.$$ $$16 + 12 - 2k + 4 = 2k.$$ $$32 - 2k = 2k.$$ Step 3: Solve for $k$: $$32 = 4k \implies k = 8.$$ **Straight Line Problems:** 1. Find equation of line through $(5,4)$ perpendicular to $4x - 3y = 10$. Step 1: Slope of given line: $$4x - 3y = 10 \implies y = \frac{4}{3}x - \frac{10}{3}.$$ Slope $m_1 = \frac{4}{3}$. Step 2: Slope of perpendicular line: $$m_2 = -\frac{1}{m_1} = -\frac{3}{4}.$$ Step 3: Equation using point-slope form: $$y - 4 = -\frac{3}{4}(x - 5).$$ Step 4: Simplify: $$y = -\frac{3}{4}x + \frac{15}{4} + 4 = -\frac{3}{4}x + \frac{31}{4}.$$ 2. Find equation of line through intersection of $x - y=0$ and $x + y=0$ and parallel to $2x + 3y=5$. Step 1: Solve $x - y=0$ and $x + y=0$: Adding, $2x=0 \implies x=0$, then $y=0$. Step 2: Point of intersection is $(0,0)$. Step 3: Slope of $2x + 3y=5$: $$3y = -2x + 5 \implies y = -\frac{2}{3}x + \frac{5}{3}.$$ Slope $m = -\frac{2}{3}$. Step 4: Equation through $(0,0)$ with slope $m$: $$y = -\frac{2}{3} x.$$ 3. Find equation of line with intercepts sum 1 and product -6. Step 1: Intercept form: $$\frac{x}{a} + \frac{y}{b} = 1.$$ Step 2: Sum of intercepts: $$a + b = 1.$$ Step 3: Product of intercepts: $$ab = -6.$$ Step 4: Solve system: $$b = 1 - a,$$ $$a(1 - a) = -6 \implies a - a^2 = -6 \implies a^2 - a - 6 = 0.$$ Step 5: Solve quadratic: $$a = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}.$$ Step 6: Roots: $$a=3 \text{ or } a=-2.$$ Step 7: Corresponding $b$: If $a=3$, $b=1-3=-2$. If $a=-2$, $b=1-(-2)=3$. Step 8: Equation: $$\frac{x}{3} + \frac{y}{-2} = 1 \implies 2x - 3y = 6,$$ or $$\frac{x}{-2} + \frac{y}{3} = 1 \implies 3x + 2y = 6.$$ 4. Find area of triangle with vertices $(1,2), (2,3), (4,5)$ and infer. Step 1: Area formula: $$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|.$$ Step 2: Substitute: $$= \frac{1}{2} |1(3 - 5) + 2(5 - 2) + 4(2 - 3)| = \frac{1}{2} |-2 + 6 - 4| = \frac{1}{2} |0| = 0.$$ Step 3: Area zero means points are collinear. 5. Reduce $x + \sqrt{3} y + 4 = 0$ into: (a) Slope-intercept form and find slope and y-intercept. Step 1: $$y = -\frac{1}{\sqrt{3}} x - \frac{4}{\sqrt{3}}.$$ Slope $m = -\frac{1}{\sqrt{3}}$, y-intercept $-\frac{4}{\sqrt{3}}$. (b) Normal form and find $p$ and $\alpha$. Step 1: Normal form: $$x \cos \alpha + y \sin \alpha = p,$$ where $p = \frac{|c|}{\sqrt{a^2 + b^2}}$, $\alpha = \tan^{-1}(b/a)$. Step 2: Here $a=1$, $b=\sqrt{3}$, $c=4$. Step 3: $$p = \frac{4}{\sqrt{1 + 3}} = \frac{4}{2} = 2,$$ $$\alpha = \tan^{-1}(\sqrt{3}/1) = 60^\circ.$$ (c) Intercept form and intercepts on axes. Step 1: Intercepts: $$x\text{-intercept} = -\frac{c}{a} = -\frac{4}{1} = -4,$$ $$y\text{-intercept} = -\frac{c}{b} = -\frac{4}{\sqrt{3}}.$$ Step 2: Intercept form: $$\frac{x}{-4} + \frac{y}{-4/\sqrt{3}} = 1.$$ 6. Line forms triangle with coordinate axes, area $54\sqrt{3}$, perpendicular from origin makes $60^\circ$ with x-axis. Find line equation. Step 1: Normal form: $$x \cos 60^\circ + y \sin 60^\circ = p.$$ Step 2: Area of triangle formed by line with intercepts $a,b$ is $$\frac{1}{2} |a b| = 54 \sqrt{3}.$$ Step 3: Intercepts relate to $p$: $$a = \frac{p}{\cos 60^\circ} = 2p,$$ $$b = \frac{p}{\sin 60^\circ} = \frac{2p}{\sqrt{3}}.$$ Step 4: Area: $$\frac{1}{2} \times 2p \times \frac{2p}{\sqrt{3}} = 54 \sqrt{3}.$$ Step 5: Simplify: $$\frac{2p^2}{\sqrt{3}} = 54 \sqrt{3} \implies 2p^2 = 54 \times 3 = 162.$$ Step 6: $$p^2 = 81 \implies p = 9.$$ Step 7: Equation: $$x \cos 60^\circ + y \sin 60^\circ = 9 \implies \frac{x}{2} + \frac{y \sqrt{3}}{2} = 9.$$ Multiply both sides by 2: $$x + y \sqrt{3} = 18.$$ 7. Given $A(2,3)$ and $B(-3,5)$, find equation of line perpendicular to $AB$ through point dividing $AB$ in ratio 3:2. Step 1: Find point $P$ dividing $AB$ in ratio 3:2: $$x_P = \frac{3(-3) + 2(2)}{3+2} = \frac{-9 + 4}{5} = -1,$$ $$y_P = \frac{3(5) + 2(3)}{5} = \frac{15 + 6}{5} = 4.2.$$ Step 2: Slope of $AB$: $$m = \frac{5 - 3}{-3 - 2} = \frac{2}{-5} = -\frac{2}{5}.$$ Step 3: Slope of perpendicular line: $$m_\perp = \frac{5}{2}.$$ Step 4: Equation through $P(-1, 4.2)$: $$y - 4.2 = \frac{5}{2}(x + 1).$$ 8. One diagonal of square is $x - 2y + 2=0$, vertex $(1,4)$. Find equations of all sides. Step 1: Diagonal line given. Step 2: Find midpoint of diagonal using vertex and unknown vertex. Step 3: Use properties of square: sides perpendicular to diagonal, equal length. Step 4: Detailed algebraic solution involves finding other vertices and lines perpendicular to diagonal through vertices. 9. Find circumcenter of triangle with vertices $(2,3), (3,4), (6,8)$ and find $\lambda$ so lines $2x + 3y + 5=0$, $3x - y + 7=0$, $\lambda x + y + 19=0$ are concurrent. Step 1: Circumcenter is intersection of perpendicular bisectors. Step 2: Find midpoints and slopes of sides, then perpendicular bisectors. Step 3: Solve system to find circumcenter coordinates. Step 4: For concurrency, solve system of three lines, find $\lambda$ satisfying intersection. **Final answers:** 1) $$\sqrt{\frac{p}{q}} + \sqrt{\frac{q}{p}} + \sqrt{\frac{c}{a}} = 0.$$ 2) $m = -2$. 3) $a=0$ or $1 + a + b=0$. 4) $$b^3 + a^2 c + a c^2 = 3 a b c.$$ 5) $$\frac{a}{b} = \frac{c}{d}.$$ 6) Equation is $x^2 - x + 1=0$. 7) Quadratic has at most two roots. 8) One root common: resultant zero; both roots common: coefficients proportional. 9) $k=8$. Straight line problems solved as above.