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,
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both of which are mixed surds, where b is a positive rational number which is not a square number of any rational number.
“WBBSE Class 10 Maths Quadratic Surd Exercise 9.1 solutions”
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
“West Bengal Board Class 10 Maths Chapter 9 Quadratic Surd Exercise 9.1 solutions”
√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.
“WBBSE Class 10 Quadratic Surd Exercise 9.1 solutions explained”
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
“WBBSE Class 10 Maths Exercise 9.1 Quadratic Surd problem solutions”
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
“Class 10 WBBSE Maths Exercise 9.1 Quadratic Surd step-by-step solutions”
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.
“WBBSE Class 10 Chapter 9 Quadratic Surd Exercise 9.1 solution guide”
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
“West Bengal Board Class 10 Maths Exercise 9.1 Quadratic Surd solutions”
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
“Class 10 WBBSE Maths Exercise 9.1 solutions for Quadratic Surd”
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.