Question

the ratio of the am and GM of two positive numbers a and b is M:N to show that a:b is equal to (m+√(m^2-n^2) :m -√(m^2 - n^2)
playing solve and get 50points ฅ'ω'ฅ ฅ'ω'ฅ ฅ'ω'ฅ​

Answer 1

ANSWER:-

Given:

The ratio of the A.M. & G.M. of two positive numbers a & b is M:N.

To prove:

Show that a:b equal to;

$a \ratio b = (m + \sqrt{ {m}^{2} - {n}^{2} } ) \ratio (m - \sqrt{ {m}^{2} - {n}^{2} } )$

Proof:

Let the two numbers be a & b.

A.M.= a+b/2 & G.M.= √ab

According to this question:

$\frac{a + b}{2 \sqrt{ab} } = \frac{m}{n} \\ Squaring \: both \: sides \:, we \: get;\\ \\ = > \frac{( {a + b)}^{2} }{4(ab)} = \frac{ {m}^{2} }{ {n}^{2} } \\ \\ = > ( {a + b) }^{2} = \frac{4ab {m}^{2} }{ {n}^{2} } \\ \\ = > (a + b) = \frac{2 \sqrt{ab}m }{n} ...............(1)$

Using this in the identity;

(a-b)² = (a+b)² - 4ab, we get;

$( {a - b)}^{2} = \frac{4ab {m}^{2} }{ {n}^{2} } - 4ab = \frac{4ab( {m}^{2} - {n}^{2} )}{ {n}^{2} } \\ \\ = > (a - b) = \frac{2 \sqrt{ab} \sqrt{ {m}^{2} - {n}^{2} } }{n} ..............(2)$

Adding equation (1) & (2), we get;

$2a = \frac{2 \sqrt{ab} }{n} (m + \sqrt{ {m}^{2} - {n}^{2} } ) \\ \\ = > a = \frac{ \sqrt{ab} }{n} (m + \sqrt{ {m}^{2} - {n}^{2} } )$

Substituting the value of a in eq.(1) we get;

$b = \frac{2 \sqrt{ab} }{n} m - \frac{ab}{n} (m + \sqrt{ {m}^{2} - {n}^{2} }) \\ \\ = > \frac{ \sqrt{ab} }{n} m - \frac{ \sqrt{ab} }{n} \sqrt{ {m}^{2} - {n}^{2} } \\ \\ = > \frac{ \sqrt{ab} }{n} (m - \sqrt{ {m}^{2} - {n}^{2} } ) \\ Therefore, \: \: a \ratio b \\ \\ = > \frac{a}{b} = \frac{ \frac{ \sqrt{ab} }{n} (m + \sqrt{ {m}^{2} - {n}^{2} } )}{ \frac{ \sqrt{ab} }{n} (m - \sqrt{ {m}^{2} - {n}^{2} }) } = \frac{(m + \sqrt{ {m}^{2} - {n}^{2} } )}{(m - \sqrt{ {m}^{2} - {n}^{2} } )} \\ Thus , \\ = > a \ratio b = (m + \sqrt{ {m}^{2} - {n}^{2} } ) \ratio (m - \sqrt{ {m}^{2} - {n}^{2} })$

Hence,

Proved.

Hope it helps ☺️

Answer 2

Step-by-step explanation:

thanks for the points

they have aldready given the answer

so there is no use