# Telangana SCERT Class 9 Math Solution Chapter 1 Real Numbers Exercise 1.4

## Telangana SCERT Solution Class IX (9) Math Chapter 1 Real Numbers Exercise 1.4

(1) (i) (5 + √7) (2 + √5)

= 5 x 2 + 5√5 + 2√7 + √7x√5

= 10 + 5√5 + 2√7 + √7×5

= 10 + 5√5 + 2√7 + √35 Ans.

(ii) (5 + √5) (5 – √5)

= 52 –  √52 [∵ a2 – b2 = (a + b) (a – b)]

= 25 – 5

= 20.

(iii) (√3 + √7)2

= (√3)2 + 2. √3. √7 + (√7)2 [(a + b)2 = a2 + 2ab + b2)

= 3 + 2√3.7 + 7

= 3 + 2√21 + 7

= 10 + 2√21

(iv) (√11 – √7) (√11 + √7)

= (√11)2  – (√7)2 [∵ a2 – b2 = (a + b) (a – b)]

= 11 – 7

= 4

(2) (i) 5 – √3 is an irrational number since √3 is not a perfect square.

(ii) √3 + √2 is an irrational number since √3 and √2 are not perfect square.

(iii) (√2 – 2)2

= (√2)2 – 2.√2.4 + (2)2

= 2 – 8√2 + 4

= 6 – 8√2

Therefore (√2 – 2)2 is an irrational number since √2 is not a perfect square upon simplifying the above equation.

(iv) 2√7/7√7 = 2/7

Therefore, 2√7/7√7 is a rational number since 2/7 is a rational number which is the simplified result of the above given eq and can be expressed in fraction.

(v) 2π is an irrational number since π is itself an irrational number.

(vi) 1/√3 is an irrational number since √3 is not a perfect square.

(vi) (2 + √2) (2 – √2) [∵ a2 – b2 = (a + b) (a – b)]

= (2)2 – (√2)2

= 4 – 2

= 2 Therefore, The given equation is a rational number since its simplified form is a whole number. vi) t4 = 256

or, √t4 = √256 [applying square on both sides]

or, t2 = ± 16

or, t = √16

or, t = ±4 is a rational number since 4 is a perfect square.

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

Solution: A surd is a number which is not the n’th root of any rational number where n is any positive integer greater than 1.

For example √2, √3, etc.

A irrational number is any number that cannot be expressed in fractions and has non-terminating decimal value.

ex – √3, √2, π, 2.1014423 etc.

∴ all surds which are not perfect square of any rational number and has non-terminating decimal values also it cannot be expressed in a fractions, are all irrational numbers, but all irrational numbers are not surds since not all irrational numbers needs to be the n’th root of any rational number for example π, e, etc. Hence, all surds are irrational numbers but not all irrational number are surds.         Updated: September 18, 2021 — 4:39 pm