Class 9 Maths Chapter 1 HOTS Questions - Number System

Q1. Find the value of:

√32 + √48√8 + √12

Solution: Since,

√32 = √16 × 2 = 4√2 ; √48 = √16 × 3 = 4√3

Similarly, √8 = √4 × 2 = 2√2 and √12 = √4 × 3 = 2√3

Therefore, 4√2 + 4√32√2 + 2√3 = 4(√2 + √3)2(√2 + √3) = 2

Q2. If √2 = 1.4142, then find the value of:

√ √2 − 1√2 + 1

Solution:

Rationalising the denominator of: √ √2 − 1√2 + 1

√2 − 1√2 + 1 × √2 − 1√2 − 1 = (√2 − 1)²(√2)² − (1)²

= (√2 − 1)²2 − 1 = (√2 − 1)²

⇒ √2 − 1√2 + 1 = √2 − 1

Substitute √2 = 1.4142:
1.4142 − 1 = 0.4142

Q3. Find the value of 'a' in:

3 − √53 + 2√5 = a√5 − 1911

Solution:

L.H.S: 3 − √53 + 2√5 × 3 − 2√53 − 2√5 (Rationalising the denominator)

= 9 − 6√5 − 3√5 + 10(3)² − (2√5)² (Using (a + b)(a − b) = a² − b²)

= 19 − 9√59 − 20 = 19 − 9√5−11 = 9√5 − 1911

So, L.H.S = R.H.S

i.e. 911 √5 − 1911 = a√5 − 1911

⇒ a = 911

Q4. Find the value of 'a' and 'b':

7 + √57 − √5 − 7 − √57 + √5 = a + 711 √5 b

Solution:

L.H.S: 7 + √57 − √5 − 7 − √57 + √5

= (7 + √5)² − (7 − √5)²(7 − √5)(7 + √5) = 49 + 5 + 14√5 − 49 − 5 + 14√549 − 5

= 4 × 7√544 = 7√511

So, L.H.S = R.H.S

R.H.S = a + 711 √5 b

Since, L.H.S = R.H.S

∴ 0 + 711 √5 = a + 711 √5 b

⇒ a = 0, b = 1

Q5. If a = 3 + √52 , then find the value of a² + 1a²

Solution:

a = 3 + √52 ⇒ 1a = 23 + √5

Now, a² + 1/a² = (a + 1/a)² − 2

⇒ a² + 1/a² = (3 + √5)/2 + 2/(3 + √5)1 − 2

= [(3 + √5)² + 2²]2(3 + √5) − 2

= 9 + 5 + 2 × 3 × √5 + 42 × 3 + 2 × √5 − 2

= 18 + 6√56 + 2√5 − 2

= (6(3 + √5))²(2(3 + √5))² − 2

= (6/2)²9 − 2

= 3² − 2 = 7

So, a² + 1a² = 7.