Question

using properties of proportion find the value of x​

Answer 1

Let us know the properties of proportion first :

1. Componendo:

$\quad\quad\quad\mathrm{\frac{a+b}{b}=\frac{c+d}{d}}$

2. Dividendo:

$\quad\quad\quad \mathrm{\frac{a-b}{b}=\frac{c-d}{d}}$

3. Componendo and Dividendo:

$\quad\quad\quad \mathrm{\frac{a+b}{a-b}=\frac{c+d}{c-d}}$

Step-by-step explanation :

Given, $\mathrm{\frac{a+\sqrt{a^{2}-2ax}}{a-\sqrt{a^{2}-2ax}}=b}$

$\to \mathrm{\frac{a+\sqrt{a^{2}-2ax}}{a-\sqrt{a^{2}-2ax}}=\frac{b}{1}}$

Using componendo and divideno rule, we get:

$\quad \mathrm{\frac{(a+\sqrt{a^{2}-2ax})+(a-\sqrt{a^{2}-2ax})}{(a+\sqrt{a^{2}-2ax})-(a-\sqrt{a^{2}-2ax})}=\frac{b+1}{b-1}}$

$\to \mathrm{\frac{a+\sqrt{a^{2}-2ax}+a-\sqrt{a^{2}-2ax}}{a+\sqrt{a^{2}-2ax}-a+\sqrt{a^{2}+2ax}}=\frac{b+1}{b-1}}$

$\to \mathrm{\frac{2a}{2\sqrt{a^{2}-2ax}}=\frac{b+1}{b-1}}$

$\to \mathrm{\frac{a}{\sqrt{a^{2}-2ax}}=\frac{b+1}{b-1}}$

Squaring both sides, we get:

$\quad \mathrm{\big(\frac{a}{\sqrt{a^{2}-2ax}}\big)^{2}=\big(\frac{b+1}{b-1}\big)^{2}}$

$\to \mathrm{\frac{a^{2}}{a^{2}-2ax}=\frac{b^{2}+2b+1}{b^{2}-2b+1}}$

$\to \mathrm{\frac{a^{2}-2ax}{a^{2}}=\frac{b^{2}-2b+1}{b^{2}+2b+1}}$

Using the divideno rule, we get:

$\quad \mathrm{\frac{a^{2}-2ax-a^{2}}{a^{2}}=\frac{b^{2}-2b+1-b^{2}-2b-1}{b^{2}+2b+1}}$

$\to \mathrm{\frac{-2ax}{a^{2}}=\frac{-4b}{b^{2}+2b+1}}$

$\to \mathrm{\frac{x}{a}=\frac{2b}{b^{2}+2b+1}}$

$\to \mathrm{x=\frac{2ab}{b^{2}+2b+1}}$

Therefore the required solution is

$\quad\quad\quad \bold{x=\frac{2ab}{b^{2}+2b+1}}$