Question

Using properties of proportion, solve each of the following for $x$:

(ii) $\frac{a+\sqrt{a^2-2ax}}{a-\sqrt{a^2-2ax}}=b$.

Answer 1

Calculate the value of $x$:

If $\frac{a}{b}=\frac{c}{d}$, then the property of componendo and dividendo states that $\frac{a+b}{a-b}=\frac{c+d}{c-d}$.

If $\frac{a}{b}=\frac{c}{d}$, then the property of convertendo states that $\frac{a}{a-b}=\frac{c}{c-d}$.

If $\frac{a}{b}=\frac{c}{d}$, then the property of invertendo states that $\frac{b}{a}=\frac{d}{c}$.

The given equation is equivalent to,

$$\begin{gathered} \frac{a+\sqrt{a^2-2ax}}{a-\sqrt{a^2-2ax}}=\frac{b}{1} \\ \Rightarrow \frac{a+\sqrt{a^2-2ax}+a-\sqrt{a^2-2ax}}{a+\sqrt{a^2-2ax}-(a-\sqrt{a^2-2ax})}=\frac{b+1}{b-1}(bycomponendoanddividendo) \\ \Rightarrow \frac{2a}{2\sqrt{a^2-2ax}}=\frac{b+1}{b-1} \\ \Rightarrow \frac{a}{\sqrt{a^2-2ax}}=\frac{b+1}{b-1} \\ \Rightarrow \frac{a^2}{a^2-2ax}=\frac{{(b+1)}^2}{{(b-1)}^2}(Squaringbothsides) \\ \Rightarrow \frac{a^2}{a^2-(a^2-2ax)}=\frac{{(b+1)}^2}{{{(b+1)}^2-(b-1)}^2}(byconvertendo) \\ \Rightarrow \frac{a^2}{2ax}=\frac{{(b+1)}^2}{{{(b+1)}^2-(b-1)}^2} \\ \Rightarrow \frac{a}{2x}=\frac{{(b+1)}^2}{4b}(U\sin gtheformulae{(a-b)}^2=a^2-2ab+b^{2}and{(a+b)}^2=a^2+2ab+b^2) \\ \Rightarrow \frac{2x}{a}=\frac{4b}{{(b+1)}^2}(byinvertendo) \\ \therefore x=\frac{2ab}{{(b+1)}^2} \end{gathered}$$

Final Answer: The value of $x=\frac{2ab}{{(b+1)}^2}$.