QUESTION 12 :

Solution :-

• Given : x = 3 - 2√2

• To find : i. 1/x

ii. x + 1/x

III. x² + 1/x²

i. We have x = 3 - 2√2

=> 1/x = 1/(3 - 2√2)

=> 1/x = (3 + 2√2) / (3 - 2√2)•(3 + 2√2)

=> 1/x = (3 + 2√2) / [ 3² - (2√2)² ]

=> 1/x = (3 + 2√2) / (9 - 8)

=> 1/x = (3 + 2√2) / 1

=> 1/x = 3 + 2√2

ii. We have x = 3 - 2√2 and 1/x = 3 + 2√2

=> x + 1/x = 3 - 2√2 + 3 + 2√2

=> x + 1/x = 6

iii. We have x + 1/x = 6

=> (x + 1/x)² = 6² {on squaring both sides}

=> x² + 2•x•(1/x) + (1/x)² = 36

=> x² + 2 + 1/x² = 36

=> x² + 1/x² = 36 - 2

=> x² + 1/x² = 34

××××××××××××××××××××××××××××××××××××

Question 13 :

Solution :-

We NEED to find the value of ;

2/(√5 + √3) + 1/(√3 + √2) - 3/(√5 + √2)

Now ,

• 2/(√5 + √3)

= 2(√5 - √3) / (√5 + √3)•(√5 - √3)

= 2(√5 - √3) / [ (√5)² - (√3)² ]

= 2(√5 - √3) / (5 - 3)

= 2(√5 - √3) / 2

= √5 - √3

• 1/(√3 + √2)

= (√3 - √2) / (√3 + √2)•(√3 - √2)

= (√3 - √2) / [ (√3)² - (√2)² ]

= (√3 - √2) / (3 - 2)

= (√3 - √2) / 1

= √3 - √2

• 3/(√5 + √2)

= 3(√5 - √2) / (√5 + √2)•(√5 - √2)

= 3(√5 - √2) / [ (√5)² - (√2)² ]

= 3(√5 - √2) / (5 - 2)

= 3(√5 - √2) / 3

= √5 - √2

Now ,

2/(√5 + √3) + 1/(√3 + √2) - 3/(√5 + √2)

= (√5 - √3) + (√3 - √2) - (√5 - √2)

= √5 - √3 + √3 - √2 - √5 + √2

= 0

Hence ,

2/(√5 +√3) + 1/(√3 +√2) - 3/(√5 +√2) = 0