Question

Find the values of a and b if

${ \frac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} - \frac{ \sqrt{7} + 1}{\sqrt{7} - 1} = a + b \sqrt{7} }$
​

Answer 1

Basic Concept Used :-

Concept of Rationalization :-

To multiply and divide the numerator and denominator of the fraction by opposite sign in denominator to remove the radicals from denominator.

Let's solve the problem now!!

Consider

$\rm :\longmapsto\:\dfrac{ \sqrt{7} - 1 }{ \sqrt{7} + 1}$

On rationalizing the denominator, we get

$\: \sf \: \: = \: \: \:\dfrac{ \sqrt{7} - 1 }{ \sqrt{7} + 1} \times \dfrac{ \sqrt{7} - 1 }{ \sqrt{7} - 1 }$

$\: \sf \: \: = \: \: \dfrac{ {( \sqrt{7} - 1) }^{2} }{ {( \sqrt{7}) }^{2} - {(1)}^{2} }$

$\: \sf \: \: = \: \: \dfrac{7 + 1 - 2 \sqrt{7} }{7 - 1}$

$\boxed{\red{\sf\because\:{(x - y)}^{2}= {x}^{2} +{y}^{2} -2xy\:and\:(x + y)(x - y)={x}^{2} - {y}^{2}}}$

$\: \sf \: \: = \: \: \dfrac{8 - 2 \sqrt{7} }{6}$

$\: \sf \: \: = \: \: \dfrac{4 -\sqrt{7} }{3}$

$\bf\implies \:\boxed{\red{\sf\:\dfrac{ \sqrt{7} - 1}{ \sqrt{7} + 1} = \dfrac{4 - \sqrt{7} }{3}}} - - - (1)$

Now,

Consider,

$\rm :\longmapsto\:\dfrac{ \sqrt{7} + 1 }{ \sqrt{7} - 1}$

On rationalizing the denominator, we get

$\: \sf \: \: = \: \: \:\dfrac{ \sqrt{7} + 1 }{ \sqrt{7} - 1} \times \dfrac{ \sqrt{7} + 1 }{ \sqrt{7} + 1 }$

$\: \sf \: \: = \: \: \dfrac{ {( \sqrt{7} + 1) }^{2} }{ {( \sqrt{7}) }^{2} - {(1)}^{2} }$

$\: \sf \: \: = \: \: \dfrac{7 + 1 + 2 \sqrt{7} }{7 - 1}$

$\boxed{\red{\sf\because\:{(x+y)}^{2}= {x}^{2} +{y}^{2}+2xy\:and\:(x + y)(x - y)={x}^{2} - {y}^{2}}}$

$\: \sf \: \: = \: \: \dfrac{8 + 2 \sqrt{7} }{6}$

$\: \sf \: \: = \: \: \dfrac{4+\sqrt{7} }{3}$

$\bf\implies \:\boxed{\red{\sf\:\dfrac{ \sqrt{7} + 1}{ \sqrt{7} - 1} = \dfrac{4 + \sqrt{7} }{3}}} - - - - (2)$

Now,

According to statement,

$\rm :\longmapsto\:\dfrac{ \sqrt{7} - 1}{ \sqrt{7} + 1} - \dfrac{ \sqrt{7} + 1}{ \sqrt{7} - 1} = a + b \sqrt{7}$

On substituting the values from equation (1) and (2), we get

$\rm :\longmapsto\:\dfrac{4 - \sqrt{7} }{3} - \dfrac{4 + \sqrt{7} }{3} = a + b \sqrt{7}$

$\rm :\longmapsto\:\dfrac{ \cancel4 - \sqrt{7} - \cancel4 - \sqrt{7} }{3}=a + b \sqrt{7}$

$\rm :\longmapsto\:\dfrac{ - \sqrt{7} - \sqrt{7} }{3}=a + b \sqrt{7}$

$\rm :\longmapsto\:\dfrac{ -2\sqrt{7}}{3}=a + b \sqrt{7}$

So, on comparing, we get

$\rm :\longmapsto\:a = 0 \: \: \: \: and \: \: \: \: b \: = - \dfrac{2}{3}$

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)