Number System Class 9th Himachal Board Maths
Chapter – Number System
Topic – Operations on Real Numbers
Question 1. Simplify: √12 + √27.
Solution:
√12 = √(4 × 3) = 2√3
√27 = √(9 × 3) = 3√3
Therefore,
√12 + √27 = 2√3 + 3√3
= 5√3
Answer: 5√3
Question 2. Simplify: 3√5 + 2√5 − √5.
Solution:
Since all terms contain √5, combine their coefficients.
3√5 + 2√5 − √5
= (3 + 2 − 1)√5
= 4√5
Answer: 4√5
Question 3. Multiply: (√3 + √2)(√3 − √2).
Solution:
Using the identity:
(a + b)(a − b) = a² − b²
Therefore,
(√3 + √2)(√3 − √2)
= (√3)² − (√2)²
= 3 − 2
= 1
Answer: 1
Question 4. Simplify: (2 + √3)².
Solution:
Using the identity:
(a + b)² = a² + 2ab + b²
Therefore,
(2 + √3)²
= 2² + 2 × 2 × √3 + (√3)²
= 4 + 4√3 + 3
= 7 + 4√3
Answer: 7 + 4√3
Question 5. Simplify: √8 × √18.
Solution:
Using the property:
√a × √b = √(ab)
Therefore,
√8 × √18 = √144
= 12
Answer: 12
Question 6. Simplify: √50 ÷ √2.
Solution:
Using the property:
√a ÷ √b = √(a/b), where b > 0
Therefore,
√50 ÷ √2 = √(50/2)
= √25
= 5
Answer: 5
Question 7. Rationalise the denominator: 1/(√5 + √2).
Solution:
Multiply the numerator and denominator by the conjugate, √5 − √2.
1/(√5 + √2) × (√5 − √2)/(√5 − √2)
= (√5 − √2)/[(√5)² − (√2)²]
= (√5 − √2)/(5 − 2)
= (√5 − √2)/3
Answer: (√5 − √2)/3
Question 8. Simplify: (√7 + √3)² − (√7 − √3)².
Solution:
Using the identity:
(a + b)² − (a − b)² = 4ab
Here, a = √7 and b = √3.
Therefore,
(√7 + √3)² − (√7 − √3)²
= 4 × √7 × √3
= 4√21
Answer: 4√21
Question 9. Simplify: (3 + √2)(3 − √2) + √16.
Solution:
Using the identity:
(a + b)(a − b) = a² − b²
Therefore,
(3 + √2)(3 − √2) + √16
= 3² − (√2)² + 4
= 9 − 2 + 4
= 11
Answer: 11
Question 10. Simplify: (√3 + √2)² + (√3 − √2)².
Solution:
Using the identities:
(a + b)² = a² + 2ab + b²
(a − b)² = a² − 2ab + b²
Therefore,
(√3 + √2)² = 3 + 2√6 + 2
= 5 + 2√6
Similarly,
(√3 − √2)² = 3 − 2√6 + 2
= 5 − 2√6
Adding both expressions:
(5 + 2√6) + (5 − 2√6)
= 10
Answer: 10