Question

Find a and b. please answer​

Answer 1

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

$\rm :\longmapsto\:\dfrac{4 + \sqrt{45} }{4 - \sqrt{45} } = a + b \sqrt{5}$

$\large\underline{\sf{To\:Find - }}$

$\rm :\longmapsto\:a \: and \: b$

Concept Used :-

Method of Rationalization

Identities Used :-

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

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

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

Given that

$\rm :\longmapsto\:\dfrac{4 + \sqrt{45} }{4 - \sqrt{45} } = a + b \sqrt{5}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{4 + \sqrt{3 \times 3 \times 5} }{4 - \sqrt{3 \times 3 \times 5} } = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } = a + b \sqrt{5}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:\dfrac{4 + 3 \sqrt{5} }{4 - 3 \sqrt{5} } \times \dfrac{4 + 3 \sqrt{5} }{4 + 3 \sqrt{5} } = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{(4 + 3 \sqrt{5} )^{2} }{ {4}^{2} - (3 \sqrt{5})^{2} } = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{16 + 45 + 24 \sqrt{5} }{16 - 45} = a + b \sqrt{5}$

$\rm :\longmapsto\:\dfrac{61 + 24 \sqrt{5} }{ - 29} = a + b \sqrt{5}$

$\rm :\longmapsto\: - \: \dfrac{61}{29} \: - \: \dfrac{24}{29} \sqrt{5} = a + b \sqrt{5}$

So, On comparing, we get

$\bf :\longmapsto\:a \: = - \: \dfrac{61}{29} \: \: \: and \: \: \: b = \: - \: \dfrac{24}{29}$

Additional Information :-

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)

Answer 2

$a = - \frac{61}{29} \\ \\ b = - \frac{24}{29}$

Refer the attachment for complete answer ✔️