Previous exercise – Surds – Exercise 7.3 – Class X
Surds – Exercise 7.4
I.Rationalise the denominator and simplify
- 8/√3
- 3/2√x
- √(5/2y)
- 1/2√(2a/5)
- 3√5/√6
B.
- 2/√3 + √2
- x/√x – √y
- √10/√5 + √3
- 3√5/√6 -√3
- √ab/√a – √b
C.
- √3 + √2/√3 – √2
- 5√2 – √3/5√2 – √3
- 4√3 + √2/4√3 – √2
- 3 + √6/√3 + 6
II. Simplify each of the following:
- √2/√3 – √2 + √5/√3 + √2
- √5/√5 – √3 – √3/√5 + √3
- √6/√5 + 2/√5 + √2
- 7√3/√6 – √3 – 2√5/√8 + √2
- √21/√3 + √7 + 2√5/√21 + √5
- If x = 2√6 + 5 find x + 1/x
III. Solve for x
(i) 3x-4/√3x + 2 = 2 + √3x – 2/2
(ii) x – 1/√x + 1 = 4 + √x – 1/2
Surds – Exercise 7.4 – Solutions:
I. Rationalise the denominator and simplify
- 8/√3
Solution:
Rational factor of denominator of 8/√3 is √3
= 8/√3 x √3/√3
= 8x√3/3
= 8√3/3
- 3/2√x
Solution:
Rational factor of denominator of 3/2√x is √x
=3/2√x x √x/√x
= 3x√x/2x
= 3√x/2x
- √(5/2y)
Solution:
Rational factor of denominator of √(5/2y) is √2y
= √(5/2y) x √(2y)/√(2y)
= √5x√(2y)/√2(y) x √(2y)
= √(10y)/2y
- 1/2√(2a/5)
Solution:
Rational factor of denominator of 1/2√(2a/5)is √5
= 1/2 x√(2a)/√5)x √5/√5
=1/2 x √(10a)/5
= √(10a)/10
- 3√5/√6
Solution:
Rational factor of denominator of 3√5/√6 is √6
= 3√5/√6 x √6/√6
= 3√5x√6/√6x√6
= 3√30/6
= √30/2
B.
- 2/√3 + √2
Solution:
In 2/√3 + √2 denominator is √3 + √2.
Rational factor of √3 + √2 is √3 – √2
2/√3 + √2 = 2/√3 + √2 x √3 – √2/√3 – √2
= 2(√3 – √2)/(√3 – √2)(√3 – √2)
= 2(√3 – √2)/3 – 2
= 2(√3 – √2)/1
= 2(√3 – √2)
- x/√x – √y
Solution:
In x/√x – √y denominator is √x – √y .
Rational factor of √x – √y is √x + √y.
x/√x – √y = x/√x – √y x √x + √y/√x + √y
= x(√x + √y)/(√x – √y)( √x + √y)
= x(√x + √y)/x – y
- √10/√5 + √3
Solution:
In √10/√5 + √3 denominator is √5 + √3.
Rational factor of √5 + √3 is √5 – √3.
√10/√5 + √3 = √10/√5 + √3 x √5 – √3/√5 – √3
= √10(√5 – √3)/(√5 + √3)(√5 – √3)
= √10(√5 – √3)/5 – 3
= √10(√5 – √3)/2
- 3√5/√6 -√3
Solution:
In 3√5/√6 -√3 denominator is √6 -√3.
Rational factor of √6 -√3 is √6 +√3.
3√5/√6 -√3 = 3√5/√6 -√3 x √6 +√3/√6 +√3
= 3√5(√6 +√3)/6 -3
= 3√5(√6 +√3)/3
= √5(√6 +√3)
- √(ab)/√a – √b
Solution:
In √(ab)/√a – √b denominator is √a – √b
Rational factor of √a – √b is √a + √b.
√(ab)/√a – √b = √(ab)/√a – √b x √a + √b/√a + √b
= √(ab)(√a + √b)/a – b
C.
- √3 + √2/√3 – √2
Solution:
In √3 + √2/√3 – √2 denominator is √3 – √2
Rational factor of √3 – √2 is √3 + √2
√3 + √2/√3 – √2 = √3 + √2/√3 – √2 x √3 + √2/√3 + √2
= (√3 + √2)(√3 + √2)/3 – 2
= (√3 + √2) (√3 + √2)
=(√3 + √2)2
= 3 + 2√2√3 + 2
= 5 + 2√6
- 5√2 – √3/3√2 – √5
Solution:
In 5√2 – √3/3√2 – √5 denominator is 3√2 – √5
Rational factor of 3√2 – √5 is 3√2 + √5
5√2 – √3/3√2 – √5 = 5√2 – √3/3√2 – √5 x 3√2 + √3/3√2 + √5
= (5√2 – √3)( 3√2 + √5)/9×2 – 5
=(5√2 – √3)( 3√2 + √5)/18 – 5
=(5√2 – √3)(3√2 + √5)/13
= (5√2(3√2 + √5) – √3)(3√2 + √5)/13
= 15×2 + 5x√10 – 3√6 – √15/13
- 4√3 + √2/√3 + √2
Solution:
In 4√3 + √2/√3 + √2 denominator is √3 + √2
Rational factor of √3 + √2 is √3 – √2
4√3 + √2/√3 + √2 = 4√3 + √2/√3 – √2 x √3 – √2/√3 – √2
= (4√3 + √2)(√3 – √2)/3 – 2
=(4√3 + √2)(√3 – √2)
= 4√3(√3 – √2) + √2(√3 – √2)
= 4×3 – 4√6 + √6 – 2
= 12 – 2 +(-4 + 1) √6
= 10 – 3√6
- 3 + √6/√3 + 6
Solution:
In 3 + √6/√3 + 6 denominator is √3 + 6
Rational factor of √3 + 6 is √3 – 6
3 + √6/√3 + 6 = 3 + √6/√3 + 6 x (√3 – 6)/( √3 – 6)
= (3 + √6) (√3 – 6)/3 – 36
= (3 + √6) (√3 – 6)/-33
= 3(√3 – 6) + √6(√3 – 6)/-33
= 3√3 – 18 + √18 – 6√6/-33
= -(√3 – 6 + √2 – 2√6/11)
II. Simplify each of the following:
- √2/√3 – √2 + √5/√3 + √2
Solution:
Rational factor of √3 – √2 is √3 + √2 and rational factor of √3 + √2 is √3 – √2
⸫√2/√3 – √2 + √5/√3 + √2 = √2/√3 – √2 x √3 + √2 /√3 + √2 + √5/√3 + √2 x √3 – √2/√3 – √2
= √2(√3 + √2)/3 – 2 + √5(√3 – √2)/3 – 2
=[√2(√3 + √2]+[ √5(√3 – √2)]
= √6 + 2 + √15 – √10
- √5/√5 – √3 – √3/√5 + √3
Solution:
Rational factor of √5 – √3 is √5 + √3 and rational factor of √5 + √3 is √5 – √3
⸫√5/√5 – √3 – √3/√5 + √3 = √5/√5 – √3 x √5 + √3/√5 + √3 – √3/√5 + √3 x √5 – √3/√5 – √3
= √5(√5 + √3)/5 – 3 – √3(√5 – √3)/5 – 3
= 5 + √15/2 – √15 – 3/2
= 5 + √15 – √15 + 3/2
= 8/2
= 4
- √6/√5 + 2/√5 + √2
Solution:
Rational factor of √5 is √5 and rational factor of √5 + √2 is √5 – √2
√6/√5 + 2/√5 + √2 = √6/√5 x √5/√5 + 2/√5 + √2 x√5 – √2/√5 – √2
= √30/5 + 2/√5 + √2 x √5 – √2/√5 – √2
= √30/5 + 2(√5 – √2)/5 – 2
= √30/5 + 2√5 – 2√2/3
= 3√30 + 10√5 – 10√2/15
= 3√30 + 10(√5 – √2)/15
- 7√3/√6 – √3 – 2√5/√8 + √2
Solution:
Rational factor of √6 – √3 is √6 + √3 and rational factor of √8 + √2 is √8 – √2
7√3/√6 – √3 – 2√5/√8 + √2 = 7√3/√6 – √3 x √6 + √3/√6 + √3 – 2√5/√8 + √2 x √8 – √2/√8 – √2
= 7√3(√6 + √3)/6-3 – 2√5(√8 – √2)/8-2
= 7√18 + 7√9/3 – 2√40 – 2√10/6
= 7√18 + 7√9/3 – √40 – √10/3
= 7√18 + 7√9 – √40 + √10/3
= 7×3√2 + 7×3 – 2√10 + √10/3
= 21√2 + 21 – 2√10 + √10/3
= 21(√2 + 1) – 2√10 + √10/3
= 21(√2 + 1) – √10/3
- √21/√3 + √7 + 2√5/√21 + √5
Solution:
Rational factor of √3 + √7 is √3 – √7 and rational factor of √21 + √5 is √21 – √5
√21/√3 + √7 + 2√5/√21 + √5 = √21/√3 + √7 x √3 – √7/√3 – √7+ 2√5/√21 + √5 x √21 – √5/√21 – √5
= √21 x (√3 – √7)/3 – 7 + 2√5(√21 – √5)/21 – 5
= √63 – √147/-4 + √105 – 5/8
= -2(√63 – √147) + √105 – 5/8
= -2√63 + 2√147 + √105 – 5/8
= -2√(7×9) + 2√(49×3) + √105 – 5/8
= -6√7 + 14√3 + √105 – 5/8
- If x = 2√6 + 5 find x + 1/x
Solution:
We have, x = 2√6 + 5.
We have to find, x + 1/x = (2√6 + 5) + 1/2√6 + 5
Rational factor of 2√6 + 5 is 2√6 – 5
(2√6 + 5) + 1/2√6 + 5 = (2√6 + 5) + 1/2√6 + 5 x 2√6 – 5/2√6 – 5
= (2√6 + 5) + 2√6 – 5/4×6 – 25
= (2√6 + 5) + 2√6 – 5/-1
= –(2√6 + 5) + 2√6 – 5/-1
=[ – 2√6 – 5 + 2√6 – 5]/-1
= -10/-1
= 10
III. Solve for x
(i) 3x-4/√(3x) + 2 = 2 + √(3x) – 2/2
Solution:
3x-4/√(3x) + 2 = 2 + √(3x) – 2/2
3x-4/√(3x) + 2 = 4+ √(3x) – 2/2
(3x – 4)2 = (4+ √(3x) – 2)( √(3x) + 2)
6x – 8 = √(3x)(4+ √(3x) – 2)+2(4+ √(3x) – 2)
6x – 8 = 4√(3x) + 3x – 2√(3x) + 8 + 2√(3x) – 4
6x – 8 = (4 – 2 + 2)√(3x) + 3x + 8 – 4
6x – 8 = 4√(3x) + 3x + 4
4√(3x) + 3x – 6x + 4 + 8 = 0
4√(3x) – 3x + 12 = 0
3x – 4√(3x) – 12 = 0
[√(3x)]2 – 6√(3x) + 2√(3x) – 12 = 0
√(3x)[ √(3x) – 6)]+2[√(3x) – 6] = 0
(√(3x) – 6)(√(3x) + 2) = 0
√(3x) = 6 or √(3x) = -2
If √(3x) = 6 , taking square on both the sides,
3x = 36
x = 36/3 = 12
Therefore, x = 12
(ii) x – 1/√x + 1 = 4 + √x – 1/2
Solution:
x – 1/√x + 1 = 8+ √x – 1/2
2(x – 1) = (8 + √x – 1)( √x + 1)
2x – 2 = (√x + 7)( √x + 1)
2x – 2 = √x (√x + 1)+ 7(√x + 1)
2x – 2 = x + √x + 7√x + 7
2x – 2 = 8√x + x + 7
2x – x – 8√x – 2 – 7 = 0
x – 8√x – 9 = 0
(√x)2 – 8√x – 9 = 0
(√x)2 – 9√x + √x – 9 = 0
√x(√x – 9) +1(√x – 9) = 0
(√x + 1)( √x – 9) = 0
√x = -1 or √x = 9
For √x = 9 then, x = 81
Therefore x = 81.
