Question

Answer this question correctly ✌.

Do one of them ✅​

Answer 1

Answer:

here x = √5 +2

we have to fine find

$( \frac{x + 1}{x} )^{2}$

put x = √5 + 2

$(\frac{ \sqrt{5} + 2 + 1 }{ \sqrt{5} + 2 } ) ^{2}$

$( \frac{ \sqrt{5} + 3 }{ \sqrt{5} + 2} ) ^{2}$

using identity (a+b)² = a² + b² + 2ab

$\frac{5 + 9 + 6 \sqrt{5} }{5 + 4 + 4 \sqrt{5} }$

$\frac{14 + 6 \sqrt{5} }{9 + 4 \sqrt{5} }$

on rationalising

$\frac{14 + 6 \sqrt{5} }{9 + 4 \sqrt{5} } \times \: \frac{9 - 4 \sqrt{5} }{9 - 4 \sqrt{5} }$

$\frac{126 - 56 \sqrt{5} + 54 \sqrt{5} - 120}{81 + 80}$

$\frac{6 - 2 \sqrt{5} }{161}$

Step-by-step explanation:

hope it helps...

Answer 2

Answer :

A) (x + 1/x)² = 20

B) a = 27/22 , b = 1/22

Solution :

• A) If x = √5 + 2 , then (x + 1/x)² = ?

We have ,

x = √5 + 2

Thus ,

1/x = 1/(√5 + 2)

Now ,

Rationalising the denominator of the term in RHS , we get ;

=> 1/x = (√5 - 2)/(√5 + 2)(√5 + 2)

=> 1/x = (√5 - 2)/[(√5)² - 2²]

=> 1/x = (√5 - 2)/(5 - 4)

=> 1/x = (√5 - 2)/1

=> 1/x = √5 - 2

Now ,

=> (x + 1/x)² = [(√5 + 2) + (√5 - 2)]²

=> (x + 1/x)² = (2√5)²

=> (x + 1/x)² = 20

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

We have ;

(7 + √5)/(7 - √5) = a + 7√5b

Now ,

Rationalising the denominator of the term in LHS , we get ;

=> (7 + √5)²/(7 - √5)(7 + √5) = a + 7√5b

=> [7² + (√5)² + 2•7•√5]/[7² - (√5)²] = a + 7√5b

=> (49 + 5 + 14√5)/(49 - 5) = a + 7√5b

=> (54 + 14√5)/44 = a + 7√5b

=> 54/44 + 14√5/44 = a + 7√5b

=> 27/22 + 7√5/22 = a + 7√5b

=> 27/22 + 7√5•(1/22) = a + 7√5b

Now ,

Comparing both the sides , we get ;

a = 27/22 , b = 1/22