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Mathematics

Using properties of proportion, find x from the following equations :

(i)2−x+2+x2−x−2+x=3(ii)x+4+x−10x+4+x−10=52(iii)1+x+1−x1+x−1−x=ab(iv)5x+2x−65x−2x−6=4(v)a+x+a−xa+x−a−x=cd(vi)a+a2−2axa−a2−2ax=b\begin{matrix} \text{(i)} & \dfrac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = 3 \\[0.5em] \text{(ii)} & \dfrac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10}} = \dfrac{5}{2} \\[0.5em] \text{(iii)} & \dfrac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \dfrac{a}{b} \\[0.5em] \text{(iv)} & \dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4 \\[0.5em] \text{(v)} & \dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \dfrac{c}{d} \\[0.5em] \text{(vi)} & \dfrac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = b \end{matrix}

Ratio Proportion

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Answer

(i) Given,

2−x+2+x2−x−2+x=31\dfrac{\sqrt{2 - x} + \sqrt{2 + x}}{\sqrt{2 - x} - \sqrt{2 + x}} = \dfrac{3}{1}

Applying componendo and dividendo,

⇒2−x+2+x+2−x−2+x2−x+2+x−2−x+2+x=3+13−1⇒22−x22+x=42⇒2−x2+x=21\Rightarrow\dfrac{\sqrt{2 - x} + \sqrt{2 + x} + \sqrt{2 - x} - \sqrt{2 + x}}{\sqrt{2 - x} + \sqrt{2 + x} - \sqrt{2 - x} + \sqrt{2 + x}} = \dfrac{3 + 1}{3 - 1} \\[1em] \Rightarrow \dfrac{2\sqrt{2 - x}}{2\sqrt{2 + x}} = \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{\sqrt{2 - x}}{\sqrt{2 + x}} = \dfrac{2}{1} \\[1em]

Squaring both sides we get,

⇒2−x2+x=41⇒(2−x)=4(2+x)⇒2−x=8+4x⇒5x=−6⇒x=−65.\Rightarrow \dfrac{2 - x}{2 + x} = \dfrac{4}{1} \\[0.5em] \Rightarrow (2 - x) = 4(2 + x) \\[0.5em] \Rightarrow 2 - x = 8 + 4x \\[0.5em] \Rightarrow 5x = -6 \\[0.5em] \Rightarrow x = -\dfrac{6}{5}.

Hence, the value of x = −65.-\dfrac{6}{5}.

(ii) Given,

x+4+x−10x+4−x−10=52\dfrac{\sqrt{x + 4} + \sqrt{x - 10}}{\sqrt{x + 4} - \sqrt{x - 10}} = \dfrac{5}{2}

Applying componendo and dividendo,

⇒x+4+x−10+x+4−x−10x+4+x−10−x−4+x−10=5+25−2⇒2x+42x−10=73⇒x+4x−10=73\Rightarrow\dfrac{\sqrt{x + 4} + \sqrt{x - 10} + \sqrt{x + 4} - \sqrt{x - 10}}{\sqrt{x + 4} + \sqrt{x - 10} - \sqrt{x - 4} + \sqrt{x - 10}} = \dfrac{5 + 2}{5 - 2} \\[1em] \Rightarrow \dfrac{2\sqrt{x + 4}}{2\sqrt{x - 10}} = \dfrac{7}{3} \\[1em] \Rightarrow \dfrac{\sqrt{x + 4}}{\sqrt{x - 10}} = \dfrac{7}{3}

Squaring both sides,

⇒x+4x−10=499⇒9(x+4)=49(x−10)⇒9x+36=49x−490⇒40x=526⇒x=52640⇒x=26320.\Rightarrow \dfrac{x + 4}{x - 10} = \dfrac{49}{9} \\[1em] \Rightarrow 9(x + 4) = 49(x - 10) \\[1em] \Rightarrow 9x + 36 = 49x - 490 \\[1em] \Rightarrow 40x = 526 \\[1em] \Rightarrow x = \dfrac{526}{40} \\[1em] \Rightarrow x = \dfrac{263}{20}.

Hence, the value of x = 26320.\dfrac{263}{20}.

(iii) Given,

1+x+1−x1+x−1−x=ab\dfrac{\sqrt{1 + x} + \sqrt{1 - x}}{\sqrt{1 + x} - \sqrt{1 - x}} = \dfrac{a}{b}

Applying componendo and dividendo,

⇒1+x+1−x+1+x−1−x1+x+1−x−1+x+1−x=a+ba−b⇒21+x21−x=a+ba−b⇒1+x1−x=a+ba−b\Rightarrow\dfrac{\sqrt{1 + x} + \sqrt{1 - x} + \sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x} - \sqrt{1 + x} + \sqrt{1 - x}} = \dfrac{a + b}{a - b} \\[1em] \Rightarrow \dfrac{2\sqrt{1 + x}}{2\sqrt{1 - x}} = \dfrac{a + b}{a - b} \\[1em] \Rightarrow \dfrac{\sqrt{1 + x}}{{\sqrt{1 - x}}} = \dfrac{a + b}{a - b}

Squaring both sides we get,

⇒1+x1−x=(a+b)2(a−b)2\Rightarrow \dfrac{1 + x}{1 - x} = \dfrac{(a + b)^2}{(a - b)^2} \\[0.5em]

By componendo and dividendo,

⇒1+x+1−x1+x−1+x=(a+b)2+(a−b)2(a+b)2−(a−b)2⇒22x=a2+2ab+b2+a2−2ab+b2a2+2ab+b2−a2+2ab−b2⇒22x=2a2+2b22ab+2ab⇒22x=2(a2+b2)4ab⇒1x=a2+b22ab⇒x=2aba2+b2.\Rightarrow \dfrac{1 + x + 1 - x}{1 + x - 1 + x} = \dfrac{(a + b)^2 + (a - b)^2}{(a + b)^2 - (a - b)^2} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{a^2 + 2ab + b^2 + a^2 - 2ab + b^2}{a^2 + 2ab + b^2 - a^2 + 2ab - b^2} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{2a^2 + 2b^2 }{2ab + 2ab} \\[1em] \Rightarrow \dfrac{2}{2x} = \dfrac{2(a^2 + b^2)}{4ab} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{a^2 + b^2}{2ab} \\[1em] \Rightarrow x = \dfrac{2ab}{a^2 + b^2}.

Hence, the value of x = 2aba2+b2.\dfrac{2ab}{a^2 + b^2}.

(iv) Given,

5x+2x−65x−2x−6=4.\dfrac{\sqrt{5x} + \sqrt{2x - 6}}{\sqrt{5x} - \sqrt{2x - 6}} = 4.

By componendo and dividendo,

⇒5x+2x−6+5x−2x−65x+2x−6−5x+2x−6=4+14−1⇒25x22x−6=53⇒5x2x−6=53\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}}{2\sqrt{2x - 6}} = \dfrac{5}{3} \\[1em] \Rightarrow \dfrac{\sqrt{5x}}{\sqrt{2x - 6}} = \dfrac{5}{3}

Squaring both sides,

⇒(5x2x−6)2=(53)2⇒5x2x−6=259⇒5x×9=25(2x−6)⇒45x=50x−150⇒5x=150⇒x=30.\Rightarrow \Big(\dfrac{\sqrt{5x}}{\sqrt{2x - 6}}\Big)^2 = \Big(\dfrac{5}{3}\Big)^2 \\[1em] \Rightarrow \dfrac{5x}{2x - 6} = \dfrac{25}{9} \\[1em] \Rightarrow 5x \times 9 = 25(2x - 6) \\[1em] \Rightarrow 45x = 50x - 150 \\[1em] \Rightarrow 5x = 150 \\[1em] \Rightarrow x = 30.

Hence, the value of x is 30.

(v) Given,

a+x+a−xa+x−a−x=cd\dfrac{\sqrt{a + x} + \sqrt{a - x}}{\sqrt{a + x} - \sqrt{a - x}} = \dfrac{c}{d}

By componendo and dividendo,

⇒a+x+a−x+a+x−a−xa+x+a−x−a+x+a−x=c+dc−d⇒2a+x2a−x=c+dc−d⇒a+xa−x=c+dc−d\Rightarrow \dfrac{\sqrt{a + x} + \sqrt{a - x} + \sqrt{a + x} - \sqrt{a - x}}{\sqrt{a + x} + \sqrt{a - x} - \sqrt{a + x} + \sqrt{a - x}} = \dfrac{c + d}{c - d} \\[1em] \Rightarrow \dfrac{2\sqrt{a + x}}{2\sqrt{a - x}} = \dfrac{c + d}{c - d} \\[1em] \Rightarrow \dfrac{\sqrt{a + x}}{\sqrt{a - x}} = \dfrac{c + d}{c - d}

Squaring both sides,

⇒a+xa−x=(c+dc−d)2⇒a+xa−x=c2+d2+2cdc2+d2−2cd\Rightarrow \dfrac{a + x}{a - x} = \Big(\dfrac{c + d}{c - d}\Big)^2 \\[1em] \Rightarrow \dfrac{a + x}{a - x} = \dfrac{c^2 + d^2 + 2cd}{c^2 + d^2 - 2cd} \\[1em]

Again applying componendo and dividendo,

⇒a+x+a−xa+x−a+x=c2+d2+2cd+c2+d2−2cdc2+d2+2cd−c2−d2+2cd⇒2a2x=2(c2+d2)4cd⇒ax=c2+d22cd⇒x=2acdc2+d2\Rightarrow \dfrac{a + x + a - x}{a + x - a + x} = \dfrac{c^2 + d^2 + 2cd + c^2 + d^2 - 2cd}{c^2 + d^2 + 2cd - c^2 - d^2 + 2cd} \\[1em] \Rightarrow \dfrac{2a}{2x} = \dfrac{2(c^2 + d^2)}{4cd} \\[1em] \Rightarrow \dfrac{a}{x} = \dfrac{c^2 + d^2}{2cd} \\[1em] \Rightarrow x = \dfrac{2acd}{c^2 + d^2}

Hence, the value of x is 2acdc2+d2.\dfrac{2acd}{c^2 + d^2}.

(vi) Given,

a+a2−2axa−a2−2ax=b1.\dfrac{a + \sqrt{a^2 - 2ax}}{a - \sqrt{a^2 - 2ax}} = \dfrac{b}{1}.

By componendo and dividendo,

⇒a+a2−2ax+a−a2−2axa+a2−2ax−a+a2−2ax=b+1b−1⇒2a2a2−2ax=b+1b−1⇒aa2−2ax=b+1b−1\Rightarrow \dfrac{a + \sqrt{a^2 - 2ax} + a - \sqrt{a^2 - 2ax}}{a + \sqrt{a^2 - 2ax} - a + \sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{2a}{2\sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em] \Rightarrow \dfrac{a}{\sqrt{a^2 - 2ax}} = \dfrac{b + 1}{b - 1} \\[1em]

Squaring both sides,

⇒a2a2−2ax=(b+1b−1)2⇒a2a2−2ax=b2+1+2bb2+1−2b\Rightarrow \dfrac{a^2}{a^2 - 2ax} = \Big(\dfrac{b + 1}{b - 1}\Big)^2 \\[1em] \Rightarrow \dfrac{a^2}{a^2 - 2ax} = \dfrac{b^2 + 1 + 2b}{b^2 + 1 - 2b} \\[1em]

Applying componendo and dividendo again,

⇒a2+a2−2axa2−a2+2ax=b2+1+2b+b2+1−2bb2+1+2b−b2−1+2b⇒2a(a−x)2ax=2(b2+1)4b⇒a−xx=b2+12b\Rightarrow \dfrac{a^2 + a^2 - 2ax}{a^2 - a^2 + 2ax} = \dfrac{b^2 + 1 + 2b + b^2 + 1 - 2b}{b^2 + 1 + 2b - b^2 - 1 + 2b} \\[1em] \Rightarrow \dfrac{2a(a - x)}{2ax} = \dfrac{2(b^2 + 1)}{4b} \\[1em] \Rightarrow \dfrac{a - x}{x} = \dfrac{b^2 + 1}{2b}

On cross-multiplication,

⇒2b(a−x)=x(b2+1)⇒2ab−2bx=b2x+x⇒b2x+x+2bx=2ab⇒x(b2+1+2b)=2ab⇒x(b+1)2=2ab⇒x=2ab(b+1)2.\Rightarrow 2b(a - x) = x(b^2 + 1) \\[0.5em] \Rightarrow 2ab - 2bx = b^2x + x \\[0.5em] \Rightarrow b^2x + x + 2bx = 2ab \\[0.5em] \Rightarrow x(b^2 + 1 + 2b) = 2ab \\[0.5em] \Rightarrow x(b + 1)^2 = 2ab \\[0.5em] \Rightarrow x = \dfrac{2ab}{(b + 1)^2}.

Hence, the value of x is 2ab(b+1)2.\dfrac{2ab}{(b + 1)^2}.

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