Exercise 1.4

1. Simplify the following expressions.

i. (5 + 7)(5 - 7)

Solution:

This is in the form of (a + b)(a - b) = a2 -+ b2.

(5 + 7)(5 - 7) = 52 - 72 = 2557 - 77 =

ii. (5+5)(5−5)

Solution:

5+55−5 = (255)() =

iii. 3+72

Solution:

3+72 = 3+73+7 = a2 + 2ab + b2

= 32 + 237 + 72 = + 6777 +

= + 677

iv. 11−711+7

Solution:

This is in the form of (a - b)(a + b) = a2 - b2.

(11 - 7)(11 + 7) = 112 - 72

= 1111 - 77 =

2. Classify the following numbers as rational or irrational.

i. 5 - 3

Solution:

IrrationalRational

Since 3 is irrational, subtracting it from a rational number 5 results in an IrrationalRational number.

ii. 3 + 2

Solution:

IrrationalRational

Since 2 is irrational, adding it to a rational number 3 results in an IrrationalRational number.

iii. 22

Solution:

RationalIrrational

22 = 2, which is a RationalIrrational number.

iv. 2+77

Solution:

IrrationalRational

2 and 7 are irrational. The sum of irrational numbers can be irrational. Dividing an irrational number by a rational number results in an IrrationalRational number.

v. 2π

Solution:

IrrationalRational

π is irrational. Multiplying an irrational number by a rational number results in an IrrationalRational number.

vi. 13

Solution:

IrrationalRational

3 is irrational. The reciprocal of an irrational number is irrational. Alternatively, rationalizing the denominator gives 33, which is clearly IrrationalRational.

vii. 2+22

Solution:

IrrationalRational

Irrational. Expanding the expression gives 4+42 + 2 = 6 + 42. Since 2 is irrational, 42 is irrational, and adding a rational number to an irrational number results in an IrrationalRational number

3. In the following equations, find whether variables x, y, z etc. represent rational or irrational numbers.

i. 2 × x = 7

Solution:

Given 2 × x = 7

x = i.e. RationalIrrational number.

ii. 2 × y2 = 16

Solution:

y2 =

y = 8 = 2223

IrrationalRational number.

iii. 2 × z = 0.02

Solution:

z = = 110011000

RationalIrrational number.

3iv. 2 × u4 = 174

Solution:

u4 =

u = 178141784.

IrrationalRational number.

v. 2 × w3 = 27

Solution:

w3 =

w = 32133213

IrrationalRational number.

vi. t4 = 256

Solution:

t = 256142564 =

RationalIrrational number.

4. Every surd is an irrational, but every irrational need not be a surd. Justify your answer.

Solution:

A surd is an irrational number that can be expressed as the nth root of a positive rational number.

For example, 2, 35, and 7 areare not surds.

All surds are irrational because they cannot be expressed as a simple fraction pq.

However, not all irrational numbers are surds.

A transcendental number, like π (pi) e or Euler's number, is irrational but cannotcan be expressed as the root of any algebraic equation with rational coefficients.

Therefore, π and e are irrational but not surds.

This demonstrates that while all surds are irrational, the set of irrational numbers is larger and includes numbers that are not surds.

5. Rationalise the denominators of the following:

i. 13+2

Solution:

Given 13+2

13+2 × 3−23−2

= (32) - 23) / (923 - 222)

= (3 - sqrt(2))/

ii. 17−6

Solution:

Given: 17−6

Multiply with 7+67+6

17−6 × 7+67+6

= (76 + 67) / ( - )

= 76 + 67

iii. 17

Solution:

Given: 17

Multiply by 77

17 × 77

= 77 / 77

iv. 6/(sqrt(3) - sqrt(2))

Solution:

Given: 63−2

now multiply it with 3+23+2

63−2 × 3+23+2

= 63+6263−62

6. Simplify each of the following by rationalising the denominator:

i. 6−426+42

Solution:

To rationalize the denominator, multiply both numerator and denominator by the conjugate of the denominator, which is 6−426+42:

6−426+42 × 6−426−42

We get numerator = 6−4226+422

And the denominator: 6+426−42 = 6232 - 422322 = - =

Now our expression is: 6−42246+4224

6−422 = 62 - 2642 + 422

= - 482362 +

= - 482

Therefore: 6−426+42 = 68−4824 = 17−12217+122

The simplified expression is 17 - 122.

ii. Simplify: 7−57+5

Solution:

7−57+5

= 7−57+5 × 7−57−5

= ( - 235225 + )/( - )

= ( - 235)/ = 6−356+35

iii. Simplify: 132−23

Solution:

132−23 = 132−23 × 32+2332+23

= (3222 + 2333) / ( - )

= 32+23632−236

iv. Simplify: 35−733+2

Solution:

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is 33−233+2:

35−733+2 = 35−733+2 × 33−233−2

= (915815 - 31037 - 21373 + 7252) / ( - )

= 915−310−213+72915−310+213+72 /

Simplify: 10−522. Then, find the approximate value to three decimal places.

Solution:

Rationalize the denominator: Multiply the numerator and denominator by 2:

10−522 × 22

Distribute 2 in the numerator:

(20210 - 107) /

Simplify 20: (20 = 4×5 = 25)

25−104

Approximate values: 5 ≈ 2.236.236 and 10 ≈ 3.1622.236

Substitute: 2×2.236−3.1624

= ( - 3.162)/4

= / 4

= (upto three decimal places)

Therefore, 10−522 simplifies to 25−104, and its approximate value to three decimal places is 0.328.

8. Find:

i. 6416

Solution:

6416 is equivalent to the sixth root of 64.

Since 26 = , the sixth root of 64 is .

Therefore, 6416 = 2.

ii. 3215

Solution:

3215 =

(iii) 62514

Solution:

62514 =

iv. 1632

Solution:

1632 = 16123 = 4344 =

v. 24315

Solution:

24315 =

vi. 4665616

Solution:

4665616 =

i. 3+23−2 = a + bsqrt6

Solution:

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is 3+23−2:

3+23−2 = 3+23−2 × 3+23+2

Expanding the numerator and denominator, we get:

( + 2636 + ) / ( - )

= ( + 26)/ = 5+265−26

Now, comparing this with the given form, a + bsqrt6, we can equate the rational and irrational parts:

a =

b =

Therefore, a = 5 and b = 2.

ii. 5+325−33 = a + bsqrt15

Solution:

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator, which is 25+3325−33:

sqrt5+sqrt32sqrt5−3sqrt3 = sqrt5+sqrt32sqrt5−3sqrt3 × 2sqrt5+3sqrt32sqrt5+3sqrt3

Expanding the numerator:

5+325+33 = + 315415 + 215315 + = + 515615

Expanding the denominator:

25−3325+33 = - =

So, we have: 19+515−7 = −197 - 5715

Comparing this with the given form, a+bsqrt15:

a =

b =

11. Find the square root of 11 + 230

Solution:

We want to find x and y such that x+y2 = 11 + 230.

Expand x+y2:

x + y + 2xy = 11 + 230

Equate the rational and irrational parts:

x + y =

xy =

Solve for x and y: We need two numbers that add up to 11 and multiply to 30. Those numbers are 67 and 59.

Assign values: Let x = 6 and y = 5 (or vice-versa, the order doesn't matter).

Substitute: 11+230 = 6+5+26×5 = 6−6 + 5−5

Therefore, the square root of 11 + 230 is 6+5.