Question

Find the value of a and b : 7 +√5 - 7- √5 = a + 7 b √5
7 - √5 7 +√5 11​

Answer 1

Step-by-step explanation:

√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11√5 - 7- √5 = a + 7 b √5

7 - √5 7 +√5 11

Answer 2

$\sf\huge\bold{Answer:}$

Given :

$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} } = a + \frac{7}{11} b \sqrt{5}$

To find :

Value of a and b

Solution :

Identities in use will be

$⇒(x - y)(x + y) = {x}^{2} - {y}^{2} \\ ⇒(x + y)(x + y) = {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ ⇒(x - y)(x - y) = {(x - y)}^{2} = {x}^{2} -2xy + {y}^{2}$

LHS

$\frac{7 + \sqrt{5} }{7 - \sqrt{5} } - \frac{7 - \sqrt{5} }{7 + \sqrt{5} }$

Now,on taking LCM

we get,

$= \frac{( {7 + \sqrt{5} )}^{2} - {(7 - \sqrt{5} )}^{2} }{(7 + \sqrt{5})(7 - \sqrt{5} ) }$

on Implementing identities ,we get

$= \frac{( {7)}^{2} + { (\sqrt{5} )}^{2} + 2(7)( \sqrt{5} ) - ( {7}^{2} + \sqrt{5} {}^{2} - 2 \times 7 \times \sqrt{5} ) }{ {7}^{2} - { \sqrt{5} }^{2} }$

On simplifying,we obtain

$= \frac{49 + 5 + 14 \sqrt{5} - (49 + 5 - 14 \sqrt{5}) }{49 - 5}$

On Removing Brackett

$= \frac{49 + 5 + 14 \sqrt{5} - 49 - 5 + 14 \sqrt{5} }{44} \\ = \frac{14 \sqrt{5} + 14 \sqrt{5} }{44} \\ = \frac{28 \sqrt{5} }{44} \\ = \frac{7 \sqrt{5} }{11}$

RHS

$= a + \frac{7}{11}b \sqrt{5}$

Now,taking LHS=RHS to find a and b

we get

$⇒ \frac{7 \sqrt{5} }{11} = a + \frac{7}{11}b \sqrt{5} \\ ⇒0 + \frac{7}{11} ( 1 )( \sqrt{5}) = a + \frac{7}{11} b \sqrt{5}$

On comparing LHS and RHS

we get

$⇒a = 0 \\ ⇒b = 1$