Question

solve this question
if a+b/√ab = 6
then find ratio of a and b
Using componendo and Dividendo rule.


The best explanation will be marked as brainlists
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Answer 1

$\textit{\underline{Componendo and dividendo}}$

If a : b then ( a + b ) : ( a - b ) .

This is a property of ratio and proportion which is going to be used in the above problem .

$\frac{a+b}{\sqrt{ab}}=6\\\\\textbf{Multiply both sides by \frac{1}{2} }\\\\\implies \frac{a+b}{2\sqrt{ab}}=\frac{6}{2}\\\textbf{By using the Componendo and dividendo we get :}\\\\\implies \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{6+2}{6-2}$

$\implies \frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}}=\frac{8}{4}\\\\\textbf{Use the formula a^2+2ab+b^2=(a+b)^2 }\\\\\implies \frac{(\sqrt{a}+\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})^2}=\frac{2}{1}\\\\\textbf{Taking square root both sides we get :-}\\\\\implies \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{2}}{1}$

$\textbf{Using componendo and dividendo we get :}\\\\\implies \frac{\sqrt{a}+\sqrt{b}+\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{a}+\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}\\\\\implies \frac{2\sqrt{a}}{2\sqrt{b}}=\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}\\\\\implies \frac{\sqrt{a}}{\sqrt{b}}=\frac{(\sqrt{2}+1)^2}{2-1}\\\\\implies \frac{\sqrt{a}}{\sqrt{b}}=(\sqrt{2}+1)^2$

$\textbf{Square both sides :}\\\\\implies \frac{a}{b}=(\sqrt{2}+1)^4\\\\\implies a:b=(2+1+2\sqrt{2})^2\\\\\implies a:b=(3+2\sqrt{2})^2\\\\\textbf{Square both sides :}\\\\\implies a:b=(2\sqrt{2})^2+3^2+2\times 2\sqrt{2}\times 3\\\\\implies a:b=9+8+12\sqrt{2}\\\\\implies a:b=17+12\sqrt{2}$

$\boxed{\boxed{\bf{\red{a:b=17+12\sqrt{2}}}}}$

Answer 2

Answer:bsbs

Step-by-step explanation:

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