## RS Aggarwal Class 9 Ex 1B Chapter 1

**Question 1: SIMPLIFY;**

**(i) ( 4 + 2–√ ) ( 4 – 2–√ ) **

**(ii) ( 5–√ + 3–√ ) (5–√ – 3–√)**

**(iii) ( 6 – 6–√) ( 6 + 6–√)**

**(iv) ( 5–√ – 2–√) (2–√ – 3–√)**

**(v) ( 5–√ – 3–√ )**

^{2}

**(vi) (3 – 3–√) ^{2}**

**Solution:**

(i) ( 4 +

= 4^{2} – (^{2} [ ( a + b) ( a – b) = a^{2} – b^{2}]

= 16 – 2

= 14

(ii) (

= (^{2} – (^{2}

= 5 – 3

= 2

(iii) ( 6 –

= 6^{2} – (^{2 }

= 36 – 6

= 30

(iv) (

=

=

(v) (^{2}

= (^{2} + (^{2} – 2 x ^{2} = a^{2} + b^{2} – 2ab ]

= 5 + 3 – 2

(vi) (3 – ^{2}

= 3^{2 }+ (^{2} – 2 x 3 x ^{2} = a^{2} + b^{2} – 2ab ]

= 9 + 3 – 6

= 12- 6

**Question 2: Represent 3.2−−−√ geometrically on the number line.
Solution:**

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the center and OA as the radius, draw a semicircle.

Now, draw BD

Here,

BD =

Taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD =

**Question 3: Represent 7.28−−−−√ geometrically on the number line.
Solution:**

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit.

Now, find the midpoint O of AC.

Taking O as the centre and OA as the radius, draw a semicircle.

Now, draw BD

We have:

BD =

Now, taking B as the centre and BD as the radius, draw an arc meeting AC produced at E.

Thus, we have:

BE = BD =

**Question 4: Mention the closure property, associative law , commutative law , the existence of Identity , existence of inverse of each real numbers for each of operations**

**addition****multiplication on real numbers**

**Solution: **

Addition Properties of Real Numbers:

(i) Closure property: The sum of two real numbers is always a real number.

(ii) Associative law: (a + b) + c = a + (b + c) for all real numbers a, b and c.

(iii) Commutative law: a + b = b + a for all real numbers a and b.

(iv) Existence of additive identity: 0 is called the additive identity for real numbers.

As, for every real number a, 0 + a = a + 0 = a

(v) Existence of additive inverse: For each real number a, there exists a real number ( -a) such that a + (-a) = 0 = (-a) + a. Here, a and (-a) are the additive inverse of each other.

Multiplication properties of Real Numbers:

(i) Closure Property: The product of two real numbers is always a real number.

(ii) Associative law: (ab)c = a(bc) for all real numbers a, b and c.

(iii) Commutative law: a x b = b x a for all real numbers a and b.

(iv) Existence of multiplicative identity: 1 is called the multiplicative identity for real numbers.

As for, every real number a, 1 x a = a x 1 = a

(v) Existence of multiplicative inverse: For each real number a , there exists a real number

**RATIONALISE THE DENOMINATOR OF EACH OF THE FOLLOWING.**

**Question 5: 7–√**

**Solution:**

On multiplying the numerator and denominator of the given number by

**Question 6:**

**Solution: 5√23√**

On Multiplying the numerator and denominator of the given number by

**Question 7: 12+3√**

**Solution:**

On Multiplying the numerator and denominator of the given number by 2 –

=

=

=

= 2 –

**Question 8: 15√−2**

**Solution:**

On Multiplying the numerator and denominator of the given number by

=

=

=

=

**Question 9: 15+32√**

**Solution:**

On Multiplying the numerator and denominator of the given number by 5 – 3

=

=

=

**Question 10: 16√−5√**

**Solution:**

On Multiplying the numerator and denominator of the given number by

=

=

**Question 11: 47√+3√**

**Solution: **

On Multiplying the numerator and denominator of the given number by

=

=

=

=

**Question 12: 3√−13√+1**

**Solution:**

On multiplying the numerator and denominator of the given number by

=

=

=

=

= 2 –

**Question 13: 3−22√3+22√**

**Solution:**

On multiplying the numerator and denominator of the given number by 3 – 2

=

=

= 17 – 12

**Question 14: 3√+13√−1 = a + b 3–√**

**Solution:**

We have:

Now, rationalizing the denominator of the given number by multiplying both the numerator and denominator with

=

=

=

=

= 2 +

Therefore, 2 +

So, on comparing the LHS and RHS, we get

a =2 and b = 1

**Question 15: 3+2√3−2√ = a + b 2–√**

**Solution:**

We have:

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 3 +

=

=

=

Therefore,

So, on comparing the LHS and the RHS, we get:

a =

**Question 16: 5−6√5+6√ a – b 6–√**

**Solution:**

We have:

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 5 –

=

=

=

=

Therefore,

So, on comparing the LHS and the RHS, we get:

a =

**Question 17: 5+23√7+43√ = a – b 3–√**

**Solution: **

Now, rationalizing the denominator of the given number by multiplying both the denominator and the numerator with 7 – 4

=

=

= 11 – 6

Therefore, 11 – 6

So, on comparing the LHS and the RHS, we get:

a = 11 and b = 6