Ratio & Proportion — Exercise 7(C)

If a : b = c : d, prove that (9a + 13b) : (9a − 13b) = (9c + 13d) : (9c − 13d).

Answer

Given,

$\dfrac{a}{b} = \dfrac{c}{d}$

Multiplying both sides by $\dfrac{9}{13}$,

$\dfrac{9a}{13b} = \dfrac{9c}{13d}$

Applying componendo and dividendo:

$\dfrac{9a + 13b}{9a - 13b} = \dfrac{9c + 13d}{9c - 13d}$

Hence, proved that (9a + 13b) : (9a − 13b) = (9c + 13d) : (9c − 13d).

If a : b = c : d, prove that (3a + 2b) : (3a − 2b) = (3c + 2d) : (3c − 2d).

Answer

Given,

$\dfrac{a}{b} = \dfrac{c}{d}$

Multiplying both sides by $\dfrac{3}{2}$,

$\dfrac{3a}{2b} = \dfrac{3c}{2d}$

Applying componendo and dividendo:

$\dfrac{3a + 2b}{3a - 2b} = \dfrac{3c + 2d}{3c - 2d}$

Hence, proved that (3a + 2b) : (3a − 2b) = (3c + 2d) : (3c − 2d).

If (3a + 5b) : (3a − 5b) = (3c + 5d) : (3c − 5d), prove that a : b = c : d.

Answer

Given,

$\dfrac{3a + 5b}{3a - 5b} = \dfrac{3c + 5d}{3c - 5d}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{(3a + 5b) + (3a - 5b)}{(3a + 5b) - (3a - 5b)} = \dfrac{(3c + 5d) + (3c - 5d)}{(3c + 5d) - (3c - 5d)} \\[1em] \Rightarrow \dfrac{3a + 5b + 3a - 5b}{3a + 5b - 3a + 5b} = \dfrac{3c + 5d + 3c - 5d}{3c +5d - 3c + 5d} \\[1em] \Rightarrow \dfrac{6a}{10b} = \dfrac{6c}{10d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}\\[1em]$

Hence, proved that a : b = c : d.

If (4a² + 7b²) : (4a² − 7b²) = (4c² + 7d²) : (4c² − 7d²), prove that a : b = c : d.

Answer

Given,

(4a² + 7b²) : (4a² − 7b²) = (4c² + 7d²) : (4c² − 7d²)

$\Rightarrow \dfrac{4a^2 + 7b^2}{4a^2 - 7b^2} = \dfrac{4c^2 + 7d^2}{4c^2 - 7d^2}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{4a^2 + 7b^2 + 4a^2 - 7b^2}{4a^2 + 7b^2 - (4a^2 - 7b^2)} = \dfrac{4c^2 + 7d^2 + 4c^2 - 7d^2}{4c^2 + 7d^2 - (4c^2 - 7d^2)} \\[1em] \Rightarrow \dfrac{8a^2}{4a^2 + 7b^2 - 4a^2 + 7b^2} = \dfrac{8c^2}{4c^2 + 7d^2 - 4c^2 + 7d^2} \\[1em] \Rightarrow \dfrac{8a^2}{14b^2} = \dfrac{8c^2}{14d^2} \\[1em] \Rightarrow \dfrac{a^2}{b^2} = \dfrac{c^2}{d^2} \\[1em] \Rightarrow \sqrt{\dfrac{a^2}{b^2}} = \sqrt{\dfrac{c^2}{d^2}} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.$

Hence, proved that a : b = c : d.

If (ma + nb) : (mc + nd) = (ma − nb) : (mc − nd), prove that a : b = c : d.

Answer

Given,

(ma + nb) : (mc + nd) = (ma − nb) : (mc − nd)

$\Rightarrow \dfrac{ma + nb}{mc + nd} = \dfrac{ma - nb}{mc - nd}$

Apply Alternendo,

$\dfrac{ma + nb}{ma - nb} = \dfrac{mc + nd}{mc - nd}$

Applying componendo and dividendo:

$\Rightarrow \dfrac{(ma + nb) + (ma - nb)}{(ma + nb) - (ma - nb)} = \dfrac{(mc + nd) + (mc - nd)}{(mc + nd) - (mc - nd)} \\[1em] \Rightarrow \dfrac{ma + nb + ma - nb}{ma + nb - ma + nb} = \dfrac{mc + nd + mc - nd}{mc + nd - mc + nd} \\[1em] \Rightarrow \dfrac{2ma}{2nb} = \dfrac{2mc}{2nd} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.$

Hence, proved that a : b = c : d.

If $\dfrac{5x + 6y}{5x - 6y} = \dfrac{5u + 6v}{5u - 6v}$, show that $\dfrac{x}{y} = \dfrac{u}{v}$.

Answer

Apply Componendo & Dividendo :

$\Rightarrow \dfrac{(5x + 6y) + (5x - 6y)}{(5x + 6y) - (5x - 6y)} = \dfrac{(5u + 6v) + (5u - 6v)}{(5u + 6v) - (5u - 6v)} \\[1em] \Rightarrow \dfrac{5x + 6y + 5x - 6y}{5x + 6y - 5x + 6y} = \dfrac{5u + 6v + 5u - 6v}{5u + 6v - 5u + 6v} \\[1em] \Rightarrow \dfrac{10x}{12y} = \dfrac{10u}{12v} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{u}{v}.$

Hence, proved that $\dfrac{x}{y} = \dfrac{u}{v}$.

If $x = \dfrac{6ab}{a + b}$, prove that $\Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2$.

Answer

Given,

$x = \dfrac{6ab}{a + b} \\[1em] x = \dfrac{3a \times 2b}{a + b} \\[1em] \dfrac{x}{3a} = \dfrac{2b}{a + b}$

Applying componendo and dividendo rule,

$\Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{2b + (a + b)}{2b - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{a + 3b}{2b - a - b} \\[1em] \Rightarrow \dfrac{x + 3a}{x - 3a} = \dfrac{a + 3b}{b - a} \text{ ...(1)}$

Again solving x,

$\Rightarrow x = \dfrac{6ab}{a + b} \\[1em] \Rightarrow x = \dfrac{3b \times 2a}{a + b} \\[1em] \Rightarrow \dfrac{x}{3b} = \dfrac{2a}{a + b}$

Apply the Componendo and Dividendo rule,

$\Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{2a + (a + b)}{2a - (a + b)} \\[1em] \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{3a + b}{2a - a - b} \\[1em] \Rightarrow \dfrac{x + 3b}{x - 3b} = \dfrac{3a + b}{a - b} \text{ .....(2)}$

Adding equations (1) and (2), we get :

$\Rightarrow \Big(\dfrac{x + 3a}{x - 3a}\Big) + \Big(\dfrac{x + 3b}{x - 3b}\Big) = \Big(\dfrac{a + 3b}{b - a}\Big) + \Big(\dfrac{3a + b}{a - b}\Big) \\[1em] = \Big(\dfrac{a + 3b}{b - a}\Big) + \Big(\dfrac{3a + b}{ - (b - a)}\Big) \\[1em] = \Big(\dfrac{a + 3b}{b - a}\Big) - \Big(\dfrac{3a + b}{b - a}\Big) \\[1em] = \Big(\dfrac{a + 3b - (3a + b)}{b - a}\Big) \\[1em] = \Big(\dfrac{a + 3b - 3a - b}{b - a}\Big) \\[1em] = \Big(\dfrac{2b - 2a}{b - a}\Big) \\[1em] = \Big(\dfrac{2(b - a)}{b - a}\Big) \\[1em] = 2.$

Hence, proved that $\Big(\dfrac{x + 3a}{x - 3a} + \dfrac{x + 3b}{x - 3b}\Big) = 2$.

If, $\dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91}$, prove that x = 11.

Answer

Given,

$\dfrac{x^{3} + 3x}{3x^{2} + 1} = \dfrac{341}{91}$

Solving L.H.S:

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(x^{3} + 3x) + (3x^{2} + 1)}{(x^{3} + 3x) - (3x^{2} + 1)} \\[1em] \Rightarrow \dfrac{(x^{3} + 3x + 3x^{2} + 1)}{(x^{3} + 3x - 3x^{2} - 1)} \\[1em] \Rightarrow \dfrac{(x + 1)^3}{(x - 1)^3} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3.$

Solving R.H.S:

Apply Componendo and Dividendo:

$\Rightarrow \dfrac{341 + 91}{341 - 91} \\[1em] \Rightarrow \dfrac{432}{250}.$

Equating L.H.S. and R.H.S.,

$\Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{432}{250} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{216}{125} \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \Big(\dfrac{6}{5}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big) = \Big(\dfrac{6}{5}\Big) \\[1em] \Rightarrow 5(x + 1) = 6(x - 1) \\[1em] \Rightarrow 5x + 5 = 6x - 6 \\[1em] \Rightarrow 6x - 5x = 6 + 5 \\[1em] \Rightarrow x = 11.$

Hence, proved that x = 11.

If $\dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$, prove that x = 7.

Answer

Given,

$\dfrac{\sqrt{x + 2} + \sqrt{x - 3}}{\sqrt{x + 2} - \sqrt{x - 3}} = 5$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(\sqrt{x + 2} + \sqrt{x - 3}) + (\sqrt{x + 2} - \sqrt{x - 3})}{(\sqrt{x + 2} + \sqrt{x - 3}) - (\sqrt{x + 2} - \sqrt{x - 3})} = \dfrac{5 + 1}{5 - 1}\\[1em] \Rightarrow \dfrac{2\sqrt{x + 2}}{2\sqrt{x - 3}} = \dfrac{6}{4}\\[1em] \Rightarrow \dfrac{\sqrt{x + 2}}{\sqrt{x - 3}} = \dfrac{3}{2}\\[1em] \Rightarrow \Big(\dfrac{\sqrt{x + 2}}{\sqrt{x - 3}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2$

Squaring both sides, we get :

$\Rightarrow \dfrac{x + 2}{x - 3} = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(x + 2) = 9(x - 3) \\[1em] \Rightarrow 4x + 8 = 9x - 27 \\[1em] \Rightarrow 9x - 4x = 27 + 8 \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \dfrac{35}{5} = 7.$

Hence, proved that x = 7.

If $\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}$, prove that x = 20.

Answer

Given,

$\dfrac{\sqrt{x + 5} + \sqrt{x - 16}}{\sqrt{x + 5} - \sqrt{x - 16}} = \dfrac{7}{3}$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16}) + (\sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16}) - (\sqrt{x + 5} - \sqrt{x - 16})} = \dfrac{7 + 3}{7 - 3} \\[1em] \Rightarrow \dfrac{(\sqrt{x + 5} + \sqrt{x - 16} + \sqrt{x + 5} - \sqrt{x - 16})}{(\sqrt{x + 5} + \sqrt{x - 16} - \sqrt{x + 5} + \sqrt{x - 16})} = \dfrac{10}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{x + 5}}{2\sqrt{x - 16}} = \dfrac{5}{2} \\[1em] \Rightarrow \dfrac{\sqrt{x + 5}}{\sqrt{x - 16}} = \dfrac{5}{2} \\[1em]$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{x + 5}}{\sqrt{x - 16}}\Big)^2 = \Big(\dfrac{5}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{x + 5}{x - 16}\Big) = \Big(\dfrac{25}{4}\Big) \\[1em] \Rightarrow 4(x + 5) = 25(x - 16) \\[1em] \Rightarrow 4x + 20 = 25x - 400 \\[1em] \Rightarrow 25x - 4x = 400 + 20 \\[1em] \Rightarrow 21x = 420 \\[1em] \Rightarrow x = \dfrac{420}{21} \\[1em] \Rightarrow x = 20.$

Hence, proved that x = 20.

If $\dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5$, prove that x = $\dfrac{3}{2}$.

Answer

(i) Given,

$\dfrac{\sqrt{3x} + \sqrt{2x - 1}}{\sqrt{3x} - \sqrt{2x - 1}} = 5$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{\sqrt{3x} + \sqrt{2x - 1} + \sqrt{3x} - \sqrt{2x - 1}}{\sqrt{3x} + \sqrt{2x - 1} - (\sqrt{3x} - \sqrt{2x - 1})} = \dfrac{5 + 1}{5 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{\sqrt{3x} + \sqrt{2x - 1} - \sqrt{3x} + \sqrt{2x - 1}} = \dfrac{6}{4} \\[1em] \Rightarrow \dfrac{2\sqrt{3x}}{2 \sqrt{2x - 1}} = \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{\sqrt{3x}}{ \sqrt{2x - 1}} = \dfrac{3}{2}$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{3x}}{ \sqrt{2x - 1}}\Big)^2 = \Big(\dfrac{3}{2}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{3x}{2x - 1}\Big) = \Big(\dfrac{9}{4}\Big) \\[1em] \Rightarrow 4(3x) = 9(2x - 1) \\[1em] \Rightarrow 12x = 18x - 9 \\[1em] \Rightarrow 18x - 12x = 9 \\[1em] \Rightarrow 6x = 9 \\[1em] \Rightarrow x = \dfrac{9}{6} = \dfrac{3}{2}$

Hence, proved that x = $\dfrac{3}{2}$.

Using properties of proportion, solve for x. Given that x is positive :

$\dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4$

Answer

Given,

$\dfrac{2x + \sqrt{4x^{2} - 1}}{2x - \sqrt{4x^{2} - 1}} = 4$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{2x + \sqrt{4x^{2} - 1} + 2x - \sqrt{4x^{2} - 1}}{2x + \sqrt{4x^{2} - 1} - (2x - \sqrt{4x^{2} - 1})} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{4x}{2x + \sqrt{4x^{2} - 1} - 2x + \sqrt{4x^{2} - 1}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{4x}{2\sqrt{(4x^2 - 1)}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2x}{\sqrt{4x^2 - 1}} = \dfrac{5}{3} \\[1em]$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{2x}{\sqrt{4x^2 - 1}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{4x^2}{4x^2 - 1}\Big) = \Big(\dfrac{25}{9}\Big) \\[1em] \Rightarrow 9(4x^2) = 25(4x^2 - 1) \\[1em] \Rightarrow 36x^2 = 100x^2 - 25 \\[1em] \Rightarrow 100x^2 - 36x^2 = 25 \\[1em] \Rightarrow 64x^2 = 25 \\[1em] \Rightarrow x^2 = \dfrac{25}{64} \\[1em] \Rightarrow x = \sqrt{\dfrac{25}{64}} \\[1em] \Rightarrow x = \dfrac{5}{8}$

Hence, x = $\dfrac{5}{8}$.

Using properties of proportion solve for x, given :

$\dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4$

Answer

Given,

$\dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{\sqrt{5x} + \sqrt{2x - 6} + \sqrt{5x} - \sqrt{2x - 6}}{\sqrt{5x} + \sqrt{2x - 6} - (\sqrt{5x} - \sqrt{2x - 6})} = \dfrac{4 + 1}{4 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{5x}}{\sqrt{5x} + \sqrt{2x - 6} - \sqrt{5x} + \sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{2\sqrt{5x}}{2\sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{\sqrt{5x}}{\sqrt{2x - 6}} = \dfrac{5}{3}$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{5x}}{\sqrt{2x - 6}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{5x}{2x - 6}\Big) = \Big(\dfrac{25}{9}\Big) \\[1em] \Rightarrow 9(5x) = 25(2x - 6) \\[1em] \Rightarrow 45x = 50x - 150 \\[1em] \Rightarrow 50x - 45x = 150 \\[1em] \Rightarrow 5x = 150 \\[1em] \Rightarrow x = \dfrac{150}{5} = 30$

Hence, x = 30.

Using Componendo and Dividendo solve for x :

$\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$

Answer

Given,

$\dfrac{\sqrt{2x + 2} + \sqrt{2x - 1}}{\sqrt{2x + 2} - \sqrt{2x - 1}} = 3$

Applying Componendo and Dividendo, we get :

$\Rightarrow \dfrac{\sqrt{2x + 2} + \sqrt{2x - 1} + \sqrt{2x + 2} - \sqrt{2x - 1}}{\sqrt{2x + 2} + \sqrt{2x - 1} - (\sqrt{2x + 2} - \sqrt{2x - 1})} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{\sqrt{2x + 2} + \sqrt{2x - 1} - \sqrt{2x + 2} + \sqrt{2x - 1}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{2\sqrt{2x + 2}}{2\sqrt{2x - 1}} = 2$

Squaring both sides, we get :

$\Rightarrow \Big(\dfrac{\sqrt{2x + 2}}{\sqrt{2x - 1}}\Big)^2 = 2^2 \\[1em] \Rightarrow \Big(\dfrac{2x + 2}{2x - 1}\Big) = 4 \\[1em] \Rightarrow 2x + 2 = 4(2x - 1) \\[1em] \Rightarrow 2x + 2 = 8x - 4 \\[1em] \Rightarrow 8x - 2x = 4 + 2 \\[1em] \Rightarrow 6x = 6 \\[1em] \Rightarrow x = \dfrac{6}{6} = 1.$

Hence, x = 1.

If $16\Big(\dfrac{a - x}{a + x}\Big)^{3} = \Big(\dfrac{a + x}{a - x}\Big)$, prove that x = $\dfrac{a}{3}$.

Answer

Given,

$\Rightarrow 16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big)$

Let, $r = \dfrac{a + x}{a - x} , \dfrac{1}{r} = \dfrac{a - x}{a + x}.$

Substituting value of r and $\dfrac{1}{r}$ in $16\Big(\dfrac{a - x}{a + x}\Big)^3 = \Big(\dfrac{a + x}{a - x}\Big)$, we get :

$\Rightarrow 16 \times \Big(\dfrac{1}{r}\Big)^3 = r \\[1em] \Rightarrow 16 = r^4 \\[1em] \Rightarrow r^4 = 2^4 \\[1em] \Rightarrow r = 2. \\[1em] \Rightarrow \dfrac{a + x}{a - x} = 2 \\[1em] \Rightarrow a + x = 2a - 2x \\[1em] \Rightarrow 3x = a \\[1em] \Rightarrow x = \dfrac{a}{3}.$

Hence, proved that x = $\dfrac{a}{3}$.

If $\dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27}$

(i) Find $\dfrac{a + b}{a - b}$

(ii) Hence using properties of proportion, find a : b.

Answer

(i) Solving,

$\Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{64}{27} \\[1em] \Rightarrow \dfrac{(a + b)^3}{(a - b)^3} = \dfrac{4^3}{3^3} \\[1em] \Rightarrow \Big(\dfrac{a + b}{a - b}\Big)^3 = \Big(\dfrac{4}{3}\Big)^3 \\[1em] \Rightarrow \dfrac{a + b}{a - b} = \dfrac{4}{3} \\[1em]$

Hence, $\dfrac{a + b}{a - b} = \dfrac{4}{3}.$

(ii) Solving further,

$\Rightarrow 3(a + b) = 4(a - b) \\[1em] \Rightarrow 3a + 3b = 4a - 4b \\[1em] \Rightarrow 4a - 3a = 3b + 4b \\[1em] \Rightarrow a = 7b \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{7}{1} \\[1em] \Rightarrow a : b = 7 : 1.$

Hence, a : b = 7 : 1.

If $x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$, prove that : x² - 4ax + 1 = 0.

Answer

Given,

$\Rightarrow x = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1}}{\sqrt{2a + 1} - \sqrt{2a - 1}}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - (\sqrt{2a + 1} - \sqrt{2a - 1})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1} + \sqrt{2a - 1} + \sqrt{2a + 1} - \sqrt{2a - 1}}{\sqrt{2a + 1} + \sqrt{2a - 1} - \sqrt{2a + 1} + \sqrt{2a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 1}}{2\sqrt{2a - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 1}}{\sqrt{2a - 1}}$

Squaring both sides, we get :

$\Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 1}{2a - 1} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 1}{2a - 1} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 1) = (x^2 + 1 - 2x)(2a + 1) \\[1em] \Rightarrow 2ax^2 - x^2 + 2a - 1 + 4ax - 2x = 2ax^2 + x^2 + 2a + 1 - 4ax - 2x \\[1em] \Rightarrow 2ax^2 - 2ax^2 + x^2 + x^2 + 2a - 2a + 1 + 1 - 4ax - 4ax - 2x + 2x = 0 \\[1em] \Rightarrow 2x^2 - 8ax + 2 = 0 \\[1em] \Rightarrow 2(x^2 - 4ax + 1) = 0 \\[1em] \Rightarrow x^2 - 4ax + 1 = 0.$

Hence, proved that x² - 4ax + 1 = 0.

If $x = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a}}{\sqrt{b + 3a} - \sqrt{b - 3a}}$, prove that : 3ax² - 2bx + 3a = 0.

Answer

Given,

$\Rightarrow x = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a}}{\sqrt{b + 3a} - \sqrt{b - 3a}}$

Applying componendo and dividendo,

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - (\sqrt{b + 3a} - \sqrt{b - 3a})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a} + \sqrt{b - 3a} + \sqrt{b + 3a} - \sqrt{b - 3a}}{\sqrt{b + 3a} + \sqrt{b - 3a} - \sqrt{b + 3a} + \sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{b + 3a}}{2\sqrt{b - 3a}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{b + 3a}}{\sqrt{b - 3a}}$

Squaring both sides:

$\Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{b + 3a}{b - 3a} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{b + 3a}{b - 3a} \\[1em] \Rightarrow (x^2 + 1 + 2x)(b - 3a) = (x^2 + 1 - 2x)(b + 3a) \\[1em] \Rightarrow bx^2 + b + 2bx - 3ax^2 - 3a - 6ax = bx^2 + b - 2bx + 3ax^2 + 3a - 6ax \\[1em] \Rightarrow bx^2 - bx^2 + b - b + 2bx + 2bx - 3ax^2 - 3ax^2 - 3a - 3a - 6ax + 6ax = 0 \\[1em] \Rightarrow 4bx - 6ax^2 - 6a = 0 \\[1em] \Rightarrow 6ax^2 - 4bx + 6a = 0 \\[1em] \Rightarrow 2(3ax^2 - 2bx + 3a) = 0 \\[1em] \Rightarrow 3ax^2 - 2bx + 3a = 0.$

Hence, proved that 3ax² - 2bx + 3a = 0.

If $x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}$, prove that : 3bx² - 4ax + 3b = 0.

Answer

Given,

$\Rightarrow x = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b}}{\sqrt{2a + 3b} - \sqrt{2a - 3b}}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - (\sqrt{2a + 3b} - \sqrt{2a - 3b})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b} + \sqrt{2a - 3b} + \sqrt{2a + 3b} - \sqrt{2a - 3b}}{\sqrt{2a + 3b} + \sqrt{2a - 3b} - \sqrt{2a + 3b} + \sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt{2a + 3b}}{2\sqrt{2a - 3b}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt{2a + 3b}}{\sqrt{2a - 3b}}$

Squaring both sides, we get :

$\Rightarrow \dfrac{(x + 1)^2}{(x - 1)^2} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow \dfrac{x^2 + 1 + 2x}{x^2 + 1 - 2x} = \dfrac{2a + 3b}{2a - 3b} \\[1em] \Rightarrow (x^2 + 1 + 2x)(2a - 3b) = (x^2 + 1 - 2x)(2a + 3b) \\[1em] \Rightarrow 2ax^2 + 2a + 4ax - 3bx^2 - 3b - 6bx = 2ax^2 + 2a - 4ax + 3bx^2 + 3b - 6bx \\[1em] \Rightarrow 2ax^2 - 2ax^2 + 2a - 2a + 4ax + 4ax - 3bx^2 - 3bx^2 - 3b - 3b - 6bx + 6bx = 0 \\[1em] \Rightarrow 8ax - 6bx^2 - 6b = 0 \\[1em] \Rightarrow 6bx^2 - 8ax + 6b = 0 \\[1em] \Rightarrow 2(3bx^2 - 4ax + 3b) = 0 \\[1em] \Rightarrow 3bx^2 - 4ax + 3b = 0.$

Hence, proved that 3bx² - 4ax + 3b = 0.

If $x = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}}$, prove that : x³ - 3x²m + 3x - m = 0.

Answer

Given,

$\Rightarrow x = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} + \sqrt[3]{m + 1} - \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} - (\sqrt[3]{m + 1} - \sqrt[3]{m - 1})} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} + \sqrt[3]{m + 1} - \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} + \sqrt[3]{m - 1} - \sqrt[3]{m + 1} + \sqrt[3]{m - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{2\sqrt[3]{m + 1}}{2\sqrt[3]{m - 1}} \\[1em] \Rightarrow \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1}}{\sqrt[3]{m - 1}}$

Cubing both sides, we get :

$\Rightarrow \Big(\dfrac{x + 1}{x - 1}\Big)^3 = \dfrac{m + 1}{m - 1} \\[1em] \Rightarrow (x + 1)^3 (m - 1) = (x - 1)^3 (m + 1) \\[1em] \Rightarrow (x^3 + 3x^2 + 3x + 1)(m - 1) = (x^3 - 3x^2 + 3x - 1)(m + 1) \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 = m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1 \\[1em] \Rightarrow m x^3 + 3m x^2 + 3m x + m - x^3 - 3x^2 - 3x - 1 -( m x^3 - 3m x^2 + 3m x - m + x^3 - 3x^2 + 3x - 1) = 0 \\[1em] \Rightarrow m x^3 - m x^3 + 3m x^2 + 3m x^2 + 3m x - 3m x + m + m - x^3 - x^3 - 3x^2 + 3x^2 - 3x - 3x - 1 + 1 = 0 \\[1em] \Rightarrow 6m x^2 + 2m - 2x^3 - 6x = 0 \\[1em] \Rightarrow 2(3mx^2 + m - x^3 - 3x) = 0 \\[1em] \Rightarrow 3mx^2 + m - x^3 - 3x = 0 \\[1em] \Rightarrow x^3 - 3mx^2 + 3x - m = 0.$

Hence, proved that x³ - 3x²m + 3x - m = 0.

What quantity must be added to each term of the ratio a : b to make it c : d ?

Answer

Let x be added.

$\therefore \dfrac{a + x}{b + x} = \dfrac{c}{d}$

⇒ d(a + x) = c(b + x)

⇒ ad + dx = cb + cx

⇒ cx - dx = ad - cb

⇒ x(c - d) = ad - bc

⇒ x = $\dfrac{ad - bc}{c - d}$

Hence, quantity that must be added = $\dfrac{ad - bc}{c - d}$.

If $\dfrac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \dfrac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d}$, prove that $\dfrac{a}{b} = \dfrac{c}{d}$.

Answer

Given,

$\Rightarrow \dfrac{a + 3b + 2c + 6d}{a - 3b + 2c - 6d} = \dfrac{a + 3b - 2c - 6d}{a - 3b - 2c + 6d}$

Applying componendo and dividendo, we get :

$\Rightarrow \dfrac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - (a - 3b + 2c - 6d)} = \dfrac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - (a - 3b - 2c + 6d)} \\[1em] \Rightarrow \dfrac{(a + 3b + 2c + 6d) + (a - 3b + 2c - 6d)}{(a + 3b + 2c + 6d) - a + 3b - 2c + 6d} = \dfrac{(a + 3b - 2c - 6d) + (a - 3b - 2c + 6d)}{(a + 3b - 2c - 6d) - a + 3b + 2c - 6d} \\[1em] \Rightarrow \dfrac{2a + 4c}{6b + 12d} = \dfrac{2a - 4c}{6b - 12d} \\[1em] \Rightarrow \dfrac{2(a + 2c)}{6(b + 2d)} = \dfrac{2(a - 2c)}{6(b - 2d)} \\[1em] \Rightarrow \dfrac{a + 2c}{b + 2d} = \dfrac{a - 2c}{b - 2d} \\[1em] \Rightarrow \dfrac{a + 2c}{a - 2c} = \dfrac{b + 2d}{b - 2d}$

Applying componendo and dividendo again,

$\Rightarrow \dfrac{(a + 2c) + (a - 2c)}{(a + 2c) - (a - 2c)} = \dfrac{(b + 2d) + (b - 2d)}{(b + 2d) - (b - 2d)} \\[1em] \Rightarrow \dfrac{(a + 2c) + (a - 2c)}{(a + 2c) - a + 2c} = \dfrac{(b + 2d) + (b - 2d)}{(b + 2d) - b + 2d} \\[1em] \Rightarrow \dfrac{2a}{4c} = \dfrac{2b}{4d} \\[1em] \Rightarrow \dfrac{a}{c} = \dfrac{b}{d} \\[1em] \Rightarrow \dfrac{a}{b} = \dfrac{c}{d}.$

Hence, proved that $\dfrac{a}{b} = \dfrac{c}{d}$.

If, $\dfrac{2a + 2b - 3c - 3d}{2a - 2b - 3c + 3d} = \dfrac{a + b - 4c - 4d}{a - b - 4c + 4d}$, prove that $\dfrac{a}{b} = \dfrac{c}{d}$.