Chapter 9 Quadratic surd Exercise 9.1
Question 1. I write by understanding 4 pure quadratic surds and 4 mixed quadratic surds.
Solution: 4 pure quadratic surds are √3,- √5
4 mixed quadratic surds are 2-√3; 2+ √6 , 3/2 – √10 , 3+ √5
Question 2. Are √4, √25 quadratic surds?
Solution: Apparently √4. √25 are in the form of surds but they are not surds.
Rational number, √4 = 2 and √25 = 5.
I apply Sreedhar Acharyya’s formula for solving the equation x2 – 2ax + (a2-b2) = 0.
We see that the roots are a + √b and a- √b,
Read and Learn More WBBSE Solutions For Class 10 Maths
both of which are mixed surds, where b is a positive rational number which is not a square number of any rational number.
Question 3. What type of number do we get by the addition, subtraction, multiplication, division, and square of the two numbers 8 and 12?
Solution: 8+ 12 = 20 (Integer)
8-12= -4 (Integer)
8 x 12 = 96 (Integer)
8/12 = 2/3 (Rational)
Question 4. We write similar surds in a specific place from the following quadratic surds.
√45, √80, √147, √180 and √500
Solution: √45, √80, √147, √180 and √500
√45
= √9×5
= 3√5
√45 = 3√5
√80
= √16×54
= 4√5
√80 = 4√5
√147
= √7x7x3
= 7√3
√147 = 7√3
√180
=√6x6x5
= 6√5
√180 = 6√5
√500
= √2x2x5x5x5
=2×5√5
=10√5
√500 =10√5
Question 5. Let us write the similar surds among the quadratic surds √48,
√27, √20 and √75
Solution: √48,√27, √20 and √75
√48
= √2x2x2x2x3
= 4√3
√27
= √3x3x3
= 3√3
√20
=√2×2×5
=2√5
√75
= √5x5x3
=5√3
√48. √27 √75 are similar surds.
Question 6. Let us write by calculating the value of (√12+√45) and (√2-√8) and see whether they can be expressed in pure quadratic surds.
Solution:
√2+ √8
= √2 + √2×2×2
= √2+2√2
= 3√2
√2-√8
= √2-√2x2x2
= √2-2√2
=-√2
(√2-√8) is a pure quadratic surd.
Question 7. Let us write by calculating the sum of √12,-4√3 and √3
Solution: (√12)+(-4√3) +6√3
= 2√3 −4√3 +6√3
= 4√3.
(√12)+(-4√3) +6√3 = 4√3.
Question 8. (9-2√5) + (12+7√5)
Solution: (9-2√5) + (12+7√5)
=9+12-2√5
= 21 +5√5
(9-2√5) + (12+7√5) = 21 +5√5
Question 9. I write any other two quadratic surds whose sum is a rational number.
Solution: (6+√7)+(6-√7)=6+6+ √7-√7 = 12
Question 10. Let us write the following numbers in the form of the product of rational and irrational numbers.
1. √175
Solution: √175
=√5x5x7
= 5√7
√175 = 5√7
2. 2 √112
Solution: 2√112
= 2.√4x4x7
=2×4√√7=8√7
2√112 =8√7
3. √108
Solution: √108
= √2x2x3x3x3
=2×3√3 =6√3
√108 =6√3
4. √125
Solution: √125
= √5x5x5
= 5√5
√125 = 5√5
5. 5√√119
Solution: 5√119
= 5√7×17
= 5√119
5√√119 = 5√119
Question 11. Let us show that √108-√75 = √3
Solution: √108 – √75 = √3
L.H.S
= √108 – √75
= √6x6x3 – √5x5x3
= 6√3-5√3
= √3
R.H.S.
Question 12. Let us show that √98+ √8-2√32 = √2
Solution: √98 + √8-2√32 = √2
L.H.S= √98 + √8 -2√32
= √7x7x2 + √2x2x2 -2√4x4x2
=7√2 +2√2 -2×4√2
=9√2-8√2
= √2
R.H.S.
Question 13. Let us show that 3 √48-4√75+ √192 = 0
Solution: 3√48 -4√75 + √192 =0
L.H.S.
= 3√48-4√75 + √192
= 3√4x4x3 -4√5x5x3 + √8x8x3
=3×4√3-4×5√3 +8√3
= 12√3-20√3 +8√3
=20√3-20√3
= 0
R.H.S.
Question 14. Let us simplify: √12 + 18+ √27 – √32
Solution: √12+ √18+ √27-√32
=√2x2x3 + √3x3x2 + √3×3×3 – √4x4x2
=2√3 +3√2 +3√3-4√2
=2√3 +3√3 +3√3 -4√2
=5√3 – √2
Question 15.
1. Let us write what should be added with √5+ √√3 to get the sum 2√5.
Solution: Required number = 2√5 – (√5+√3)
=2√5-√5-√3
=√5-√3
2. Let us write what should be subtracted from 7-√3 to get the sum of 25.
Solution: Required number = (7-√3)-(3+√3)
=7-√3-3-√3
=7-3-√3
=4-2√3.
3. Let us write the sum of 2+√3, √3+ √5, and 2+ √7.
Solution: Required sum = 2 + √3 + √3 + √5 +2+√7
=2+2 + √√3 + √√3 + √5 + √7
=4+2√3 +√5+√7.
4. Let us subtract (-5+3 √11) from (10+ √11) and let us write the value of the subtraction.
Solution: Required subtraction = (10-√11)-(-5+3√11)
= 10- √11+5-3√11
= 15-4√11
5. Let us subtract (5+ √2+ √7) from the sum of (-5+ √7) and (√7 + √2) and find the value of the subtraction.
Solution: Required value of subtraction = (-5+√7) + (√7+√2) – (5+√2+√7)
=-5+ √7 + √7 + √2-5-√2-√7
=-10+ √7
6. I write two quadratic surds whose sum is a rational number.
Solution: Two quadratic surds whose sum is a rational number,
5+√3:5-√3.
Question 16. Let us write by calculating the product of (3+ √7 √5) and (2√2-1)
Solution: (3+ √7-√5) x (2√2-1)
=6√2-3+2√14 – √7-2 √10-√5
Question 17. Let us write two rationalizing factors of √7.
Solution: √7 & 2√7
Question 18. Let us see what will be the rationalizing factor of (5+ √7).
Solution : (5+√7)
= (5+√7)x (5-√7)
= (5)²- (√7)²
= 25-7
=18 [ (a+b) (a-b) = a2-b2]
Again, (5+√7)x(5+7)
=(√7 +5) (√7 -5)
=(√7)²- (5)²
=-18
Question 19. Let us write two rationalizing factors of 7-√3
Solution: 7-√3
=(7+√3); (-7-√3)
Question 20. Let us see the rationalizing factors of (√11 – √6).
Solution: (√11-√6) (√11+√6)
=(√11)²-(√6)²
= 11-6
=5
Again, (√11-√6) √11-√6) = [(√11-√6) (√11+√6)]
= [11 – 6]
=-5
Question 21. Let us write two rationalizing factors of (√15+ √3)
Solution : (√15+√3)
(√15-√3): (-√15+√3)
Question 22. Let us write the conjugate guards of the following mixed and pure surds
1. 2+√3
Solution: 2+√3
=2-√3
2. 5-√2
Solution: 5-√2
=5+√2
3. √5-7
Solution: √5-7
=-√5+7
4. √11 + 6
Solution: √11 + 6
= (6-√11)
5. √5
Solution: √5
= – √5
Question 23. Let us rationalize the denominator of
1. 4√5 / 5/√3
Solution:4√5 / 5√3
=4√5.√3 / 5√3.√3
=4√3/5×3
= 4√3 / 15
4√5 / 5√3 = 4√3 / 15
2. √6/3√7
Solution: √6/3√7
= 3√7/√6
= 3√7x√6 /√6x√6
= 3√42 /6
= √42/2
√6/3√7 = √42/2
Question 24. Let us rationalize the denominator of
1. (4+2√3)+(2-√3)
Solution: 4+2√3 / 2-√3
=(4+2√3) (2+√3) /(2-√3)2+√3)
=8+4√3 +4√3 +6/(√2)²-(√3)²
=14+8√3/4-3
= 14+8√3
4+2√3 / 2-√3 = 14+8√3
2. (√5+ √3) + (√5 – √3)
Solution: √5+√3/√5-√3
=(√5+√3)(√5+√3)/(√5-√3)(√5+√3)
= 5+2√5.√3+3 /(√5)²-(√3)²
=8+2√15/ 5-3
=2(24+√15)/2
= 4+ √15.
√5+√3/√5-√3 = 4+ √15.