Question

If the sum of two numbers a and b is 3 times their GM and given that a:b= p+√q:p-√q where p and q are prime number find p+q​

Answer 1

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

Given that,

$\rm \: a : b \: = \:( p + \sqrt{q}) : (p - \sqrt{q})$

where p and q are prime numbers.

Also, given that sum of two numbers a and b is 3 times their GM.

Let assume that a > b.

$\rm \: a + b = 3 GM$

$\rm \: a + b = 3\sqrt{ab}$

can be further rewritten as

$\rm \: a + b = \dfrac{3}{2} \times 2\sqrt{ab}$

can be further rewritten as

$\rm \: \dfrac{a + b}{2 \sqrt{ab} } = \dfrac{3}{2}$

Apply Componendo and Dividendo, we get

$\rm \: \dfrac{a + b + 2 \sqrt{ab} }{a + b - 2 \sqrt{ab} } = \dfrac{3 + 2}{3 - 2}$

can be further rewritten as

$\rm \: \dfrac{ {( \sqrt{a} )}^{2} + {( \sqrt{b} )}^{2} + 2 \sqrt{ab} }{ {( \sqrt{a} )}^{2} + {( \sqrt{b} )}^{2} - 2 \sqrt{ab} } = \dfrac{5}{1}$

can be further reduced as

$\rm \: \dfrac{ {( \sqrt{a} + \sqrt{b} )}^{2} }{ {( \sqrt{a} - \sqrt{b})}^{2} } = 5$

$\rm \: \dfrac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} - \sqrt{b} } = \sqrt{5}$

can be rewritten as

$\rm \: \dfrac{ \sqrt{a} + \sqrt{b} }{ \sqrt{a} - \sqrt{b} } = \dfrac{ \sqrt{5} }{1}$

On applying Componendo and Dividendo, we get

$\rm \: \dfrac{ \sqrt{a} + \sqrt{b} + \sqrt{a} - \sqrt{b} }{ \sqrt{a} + \sqrt{b} - \sqrt{a} + \sqrt{b} } = \dfrac{ \sqrt{5} + 1}{ \sqrt{5} - 1}$

$\rm \: \dfrac{ 2\sqrt{a}}{2\sqrt{b}} = \dfrac{ \sqrt{5} + 1}{ \sqrt{5} - 1}$

$\rm \: \dfrac{\sqrt{a}}{\sqrt{b}} = \dfrac{ \sqrt{5} + 1}{ \sqrt{5} - 1}$

On squaring both sides, we get

$\rm \: \dfrac{a}{b} = \dfrac{ {( \sqrt{5} + 1)}^{2} }{ {( \sqrt{5} - 1)}^{2} }$

$\rm \: \dfrac{a}{b} = \dfrac{5 + 1 + 2 \sqrt{5} }{5 + 1 - 2 \sqrt{5} }$

$\rm \: \dfrac{a}{b} = \dfrac{6 + 2 \sqrt{5} }{6 - 2 \sqrt{5} }$

$\rm \: \dfrac{a}{b} = \dfrac{2(3 + \sqrt{5}) }{2(3 - \sqrt{5} )}$

$\rm \: \dfrac{a}{b} = \dfrac{3 + \sqrt{5}}{3 - \sqrt{5}}$

$\rm\implies \: a \: : \: b= (3 + \sqrt{5}) \: : \: (3 - \sqrt{5})$

But given that

$\rm \: a : b \: = \:( p + \sqrt{q}) : (p - \sqrt{q})$

So, on comparing we get

$\rm \: p = 3$

and

$\rm \: q = 5$

So,

$\rm\implies \:\rm \: p + q = 3 + 5 = 8$

$\rule{190pt}{2pt}$

Formulae Used :-

1. If a and b are two positive real numbers, then Geometric mean between a and b is given by

$\boxed{ \rm{ \: \: GM \: \: = \: \: \sqrt{ab} \: \: }}$

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

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

2. If a : b :: c : d, then Componendo and Dividendo means

$\boxed{ \rm{ \: \: \frac{a + b}{a - b} = \frac{c + d}{c - d} \: \: }}$