Question

If (7 + √5)/(7 - √5) = a + 7√5b , then a , b = ? ​

Answer 1

Given

• $\sf \dfrac{7 + \sqrt{5} }{7 - \sqrt{5}} = a + 7 \sqrt{5}b$

To find

• The value of a and b

Solution

$\qquad : \implies \sf \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} = a + 7 \sqrt{5} b$

$\qquad : \implies \sf \dfrac{7 + \sqrt{5} }{7 - \sqrt{5}}$

$\qquad : \implies \sf \dfrac{(7 + \sqrt{5})(7 + \sqrt{5} ) }{(7 - \sqrt{5})(7 + \sqrt{5} )}$

$\qquad : \implies \sf \dfrac{(7 + \sqrt{5})^{2} }{(7)^{2} - (\sqrt{5 )}^{2}}$

$\qquad : \pink{ identities \: to \: be\: used \mapsto}$

$\qquad : \tt{ \bullet \: \: (a + b)^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\qquad : \bullet \tt \: \: ( {a}^{2} - {b}^{2} ) = (a - b)(a + b)$

$\qquad : \implies \sf \dfrac{( {7)}^{2} + { (\sqrt{5} })^{2} + 2(7)( \sqrt{5)}}{( {7})^{2} - { (\sqrt{5} })^{2}}$

$\qquad : \implies \sf \dfrac{49 + 5 + 14 \sqrt{5} }{49 - 5}$

$\qquad : \implies \sf \dfrac{54 + 14 \sqrt{5} }{44}$

$\qquad : \implies \sf \dfrac{2(27 + 7 \sqrt{5}) }{44}$

$\qquad : \implies \sf \dfrac{ \not2(27 + 7 \sqrt{5}) }{ \cancel{44}}$

$\qquad : \implies \sf \dfrac{27 + 7 \sqrt{5} }{22}$

$\qquad : \implies \sf \dfrac{27 + 7 \sqrt{5} }{22} = a + 7 \sqrt{5} \: b$

$\qquad : \implies \sf a = \dfrac{27}{22}$

$\qquad : \implies \sf \cancel{ 7 \sqrt{5}} \: b = \dfrac{ \cancel{7 \sqrt{5} }}{22}$

$\qquad : \implies \sf b = \dfrac{1}{22}$

Answer ⟼

$\bigstar\boxed{ \underline{ \bf \red{a = \cfrac{27}{22}}}}$

$\bigstar\boxed{ \underline{ \bf \red{b = \cfrac{1}{22}}}}$

Answer 2

Answer:

Well, first you need to simplify your expression.

How ?

Simple! By rationalising.

(7+√5)/(7-√5) - (7-√5)/7+√5)

Rationalise both the terms

(7+√5)² - (7-√5)²/ 49 - 5

= 49 + 14√5 + 5 - 49 + 14√5 - 5/ 44

= 28√5/44

= 7√5/11

Now, 7√5/11 = a + (7/11) √5b

On comparison,

a = 0

b = 1

Step-by-step explanation: