Question

If
$\frac{2 \sqrt{7 } + 3 \sqrt{5} }{ \sqrt{7} + \sqrt{5} }$
=
$p \sqrt{35} + q$
Then what is the value of 2p+q​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\dfrac{2 \sqrt{7 } + 3 \sqrt{5} }{ \sqrt{7} + \sqrt{5} } = q + p \sqrt{35}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{2 \sqrt{7 } + 3 \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \times \dfrac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} - \sqrt{5} } = q + p \sqrt{35}$

We know,

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

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{14 - 2 \sqrt{35} + 3 \sqrt{35} - 15}{( \sqrt{7}) ^{2} - (\sqrt{5})^{2} } = q + p \sqrt{35}$

$\rm :\longmapsto\:\dfrac{ - 1 + \sqrt{35}}{7 - 5 } = q + p \sqrt{35}$

$\rm :\longmapsto\:\dfrac{ - 1 + \sqrt{35}}{2 } = q + p \sqrt{35}$

$\rm :\longmapsto\: - \: \dfrac{1}{2} + \dfrac{\sqrt{35}}{2 } = q + p \sqrt{35}$

So, on comparing, we get

$\purple{\rm :\longmapsto\:p = \dfrac{1}{2}}$

$\purple{\rm :\longmapsto\:q \: = \: - \: \dfrac{1}{2}}$

So,

$\purple{\rm :\longmapsto\:2p + q}$

$\rm \:  =  \: 2 \times \dfrac{1}{2} - \dfrac{1}{2}$

$\rm \:  =  \: 1 - \dfrac{1}{2}$

$\rm \:  =  \:\dfrac{1}{2}$

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More Identities to know :

➢  (a + b)² = a² + 2ab + b²

➢  (a - b)² = a² - 2ab + b²

➢  a² - b² = (a + b)(a - b)

➢  (a + b)² = (a - b)² + 4ab

➢  (a - b)² = (a + b)² - 4ab

➢  (a + b)² + (a - b)² = 2(a² + b²)

➢  (a + b)³ = a³ + b³ + 3ab(a + b)

➢  (a - b)³ = a³ - b³ - 3ab(a - b)