AP State Syllabus AP Board 9th Class Maths Solutions Chapter 1 Real Numbers Ex 1.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

Question 1.

Simple the following expressions.

i) (5 + √7) (2 + √5)

Solution:

(5 + √7) (2 + √5)

= 10 + 5√5 + 2√7 + √35

ii) (5 + √5) (5 – √5)

Solution:

(5 + √5) (5 – √5)

= 5^{2} + (√5)^{2}

= 25 – 5 = 20

(iii) (√3 + √7)^{2}

Solution:

(√3 + √7)^{2}

= (√3)^{2} + (√7)^{2} + 2(√3)(√7)

= 3 + 7 + 2√21

= 10 + 2√21

iv) (√11 – √7) (√11 + √7)

= (√11)^{2} – (√7)^{2}

= 11 – 7 = 4

Question 2.

Classify the following numbers as rational or irrational.

i) 5 – √3

ii) √3 + √2

iii) (√2 – 2)^{2}

iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)

v) 2π

vii) (2 +√2) (2 – √2)

Solution:

i) 5 – √3 – irrational

ii) √3 + √2 – irrational

iii) (√2 – 2)^{2} – irrational

iv) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\) – rational

v) 2π – Transcendental number. (not irrational)

vi) \(\frac{1}{\sqrt{3}}\)– irrational

vii) (2 +√2) (2 – √2) – rational

Question 3.

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

i) x^{2} = 7

ii) y^{2} = 16

iii) z^{2} = 0.02

iv) u^{2} = \(\frac{17}{4}\)

v) w^{2} = 27

vi) t^{4} = 256

Solution:

i) x^{2} = 7

⇒ x = √7 is an irrational number.

ii) y^{2} = 16 ⇒ y = 4 is a rational number.

iii) z^{2} = 0.02 ⇒ z = \(\sqrt{0.02}\) is an irrational number.

iv) u^{2} = \(\frac{17}{4}\) ⇒ x = \(\frac{\sqrt{17}}{2}\) is an irrational number.

v) w^{2} = 27 ⇒ w = \(3 \sqrt{3}\) an irrational number.

vi) t^{4} = 256 ⇒ t^{2} = \(\sqrt{256}\) = 16

⇒ t = \(\sqrt{16}\) = 4 is a rational number

Qeustion 4.

The ratio of circumference to the diameter of a circle c/d is represented by π. But we say that π is an irrational number. Why?

Question 5.

Rationalise the denominators of the following.

i) \(\frac{1}{3+\sqrt{2}}\)

Solution:

ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

Solution:

iii) \(\frac{1}{\sqrt{7}}\)

Solution:

iv) \(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{2}}\)

Solution:

Question 6.

Simplify each of the following by rationalising the denominator.

i) \(\frac{6-4 \sqrt{2}}{6+4 \sqrt{2}}\)

Solution:

ii) \(\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

Solution:

iii) \(\frac{1}{3 \sqrt{2}-2 \sqrt{3}}\)

Solution:

iv) \(\frac{3 \sqrt{5}-\sqrt{7}}{3 \sqrt{3}+\sqrt{2}}\)

Solution:

Question 7.

Find the value of \(\frac{\sqrt{10}-\sqrt{5}}{2 \sqrt{2}}\) upto three decimal places. (take \(\sqrt{2}\) = 1.414, \(\sqrt{3}\) = 1.732 and \(\sqrt{5}\) = 2.236).

Solution:

Question 8.

Find

i) 64^{1/6}

Solution:

= (2^{6})^{1/6}

= 6

ii) 32^{1/5}

Solution:

32^{1/5}

= (2^{5})^{1/5}

= 2

iii) 625^{1/4}

625^{1/5}

= (5^{4})^{1/4}

= 5

iv) 16^{3/2}

Solution:

16^{3/2}

= (4^{2})^{3/2}

v) 243^{2/5}

Solution:

243^{2/5}

= (3^{5})^{2/5}

vi) (46656)^{-1/6}

Solution:

Question 9.

Simplify \(\sqrt[4]{81}-8 \sqrt[3]{343}+15 \sqrt[5]{32}+\sqrt{225}\)

Solution:

Question 10.

If ‘a’ and ‘b’ are rational numbers, find the values of a and b in each of the following equations.

i) \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)

Solution:

Given that \(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\mathbf{a}+\mathbf{b} \sqrt{6}\)

Rationalising the denominator we get

Comparing 5 + 2√6 with a + b√6

We have a = 5 and b = 2

ii) \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)

Solution:

Given that \(\frac{\sqrt{5}+\sqrt{3}}{2 \sqrt{5}-3 \sqrt{3}}=a-b \sqrt{15}\)

Rationalising the denominator we get