Question

$\huge\red{\boxed{\sf QuEsTIoN}}$

The ratio of the A.M and G.M of two positive numbers a and b, is m:n. Show that

$a:b=(m + \sqrt{{ {m}^{2} } - {n}^{2}} ):(m - \sqrt{{ {m}^{2}} - {n}^{2}} )\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{A.M\,\,:\,\,G.M=m\,\,:\,\,n}$

Now,

$\sf{A.M=\dfrac{a+b}{2}\,\,\,\,\,\,and\,\,\,\,\,\,G.M=\sqrt{ab}}$

$\tt{\implies\,\dfrac{a+b}{2}\,\,:\,\,\sqrt{ab}=m\,\,:\,\,n}$

$\tt{\implies\,\dfrac{a+b}{2\sqrt{ab}}=\dfrac{m}{n}}$

$\tt{\implies\,\dfrac{(a+b)^2}{4ab}=\dfrac{m^2}{n^2}\,\,\,\,\,\,\,\,...(1)}$

$\tt{\implies\,\dfrac{a^2+b^2+2ab}{4ab}=\dfrac{m^2}{n^2}}$

Using the Dividendo rule

$\tt{\implies\,\dfrac{a^2+b^2+2ab-4ab}{4ab}=\dfrac{m^2-n^2}{n^2}}$

$\tt{\implies\,\dfrac{a^2+b^2-2ab}{4ab}=\dfrac{m^2-n^2}{n^2}}$

$\tt{\implies\,\dfrac{(a-b)^2}{4ab}=\dfrac{m^2-n^2}{n^2}\,\,\,\,\,\,\,\,...(2)}$

Divide (1) by (2)

$\tt{\implies\,\dfrac{(a+b)^2}{(a-b)^2}=\dfrac{m^2}{m^2-n^2}}$

$\tt{\implies\,\dfrac{a+b}{a-b}=\dfrac{m}{\sqrt{m^2-n^2}}}$

Applying componendo and dividendo,

$\tt{\implies\,\dfrac{a+b+a-b}{a+b-a+b}=\dfrac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}}$

$\tt{\implies\,\dfrac{2a}{2b}=\dfrac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}}$

$\tt{\implies\,\dfrac{a}{b}=\dfrac{m+\sqrt{m^2-n^2}}{m-\sqrt{m^2-n^2}}}$

$\tt{\implies\,a:b=\left(m+\sqrt{m^2-n^2}\right):\left(m-\sqrt{m^2-n^2}\right)}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

The ratio of the A.M and G.M of two positive numbers a and b, is m:n.

$\rm \implies\:\dfrac{A.M}{G.M} = \dfrac{m}{n}$

We know,

$\boxed{ \tt{ \: A.M = \frac{a + b}{2} \: }} \: \\ \\ \boxed{ \tt{ \: G.M = \sqrt{ab} \: }}$

So,

$\rm :\longmapsto\:\dfrac{a + b}{2 \sqrt{ab} } = \dfrac{m}{n}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{a + b + 2 \sqrt{ab} }{a + b - 2 \sqrt{ab} } = \dfrac{m + n}{m - n}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {( \sqrt{a} )}^{2} + {( \sqrt{b} )}^{2} + 2 \sqrt{ab} }{ {( \sqrt{a} )}^{2} + {( \sqrt{b} )}^{2} - 2 \sqrt{ab} } = \dfrac{m + n}{m - n}$

We know,

$\boxed{ \tt{ \: {x}^{2} + {y}^{2} + 2xy = {x + y)}^{2} \: }} \\ \\ \boxed{ \tt{ \: {x}^{2} + {y}^{2} - 2xy = {(x - y)}^{2} \: }}$

So, using these, we get

$\rm :\longmapsto\:\dfrac{ {( \sqrt{a} + \sqrt{b} )}^{2} }{ {( \sqrt{a} - \sqrt{b} )}^{2} } = \dfrac{m + n}{m - n}$

$\rm \implies\:\dfrac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} - \sqrt{b} } = \dfrac{ \sqrt{m + n} }{ \sqrt{m - n} }$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{ \sqrt{a} + \sqrt{b} + \sqrt{a} - \sqrt{b} }{ \sqrt{a} + \sqrt{b} - \sqrt{a} + \sqrt{b} } = \dfrac{ \sqrt{m + n} + \sqrt{m - n} }{ \sqrt{m + n} - \sqrt{m - n} }$

$\rm :\longmapsto\:\dfrac{2\sqrt{a}}{2\sqrt{b} } = \dfrac{ \sqrt{m + n} + \sqrt{m - n} }{ \sqrt{m + n} - \sqrt{m - n} }$

$\rm :\longmapsto\:\dfrac{\sqrt{a}}{\sqrt{b} } = \dfrac{ \sqrt{m + n} + \sqrt{m - n} }{ \sqrt{m + n} - \sqrt{m - n} }$

On squaring both sides, we get

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{ {[ \sqrt{m + n} + \sqrt{m - n}]}^{2} }{[ \sqrt{m + n} - \sqrt{m - n}]^{2} }$

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{m + n + m - n + 2 \sqrt{m + n} \sqrt{m - n} }{m + n + m - n - 2 \sqrt{m + n} \sqrt{m - n} }$

$\rm :\longmapsto\:\dfrac{a}{b} = \dfrac{2m + 2 \sqrt{ {m}^{2}-{n}^{2}}}{2m - 2 \sqrt{ {m}^{2}-{n}^{2}}}$

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{a}{b} = \dfrac{m + \sqrt{ {m}^{2}-{n}^{2}}}{m - \sqrt{ {m}^{2}-{n}^{2}}} \: }}$

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More to know :-

If a and b are two positive real numbers, then

$\boxed{ \tt{ \: A.M = \frac{a + b}{2} \: }} \: \\ \\ \boxed{ \tt{ \: G.M = \sqrt{ab} \: }} \\ \\ \boxed{ \tt{ \: H.M = \frac{2ab}{a + b} \: }}$

$\boxed{ \tt{ \: A.M \geqslant G.M \geqslant H.M \: }}$

$\boxed{ \tt{ \: {(G.M)}^{2} = (A.M)(H.M) \: }}$