RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities Exercise 4.1

 Exercise 4.1

Page-4.6

Question 1:

Evaluate each of the following using identities:

(i) ${\left(2x-\frac{1}{x}\right)}^2$

(ii) (2x + y) (2x − y)

(iii) (a²b - b²a)²

(iv) (a - 0.1) (a + 0.1)

(v) (1.5x² − 0.3y²) (1.5x² + 0.3y²)

Answer 1:

In the given problem, we have to evaluate expressions by using identities.

(i) Given

We shall use the identity

By applying identity we get 

Hence the value of is

(ii) We have been given

We shall use the identity

Here,,

By applying identity we get 

Hence the value ofis

(iii) The given expression is

We shall use the identity

Here

By applying identity we get 

Hence the value of is

(iv) The given expression is $\left(a+0.1\right)\left(a-0.1\right)$

We shall use the identity

Here

By applying identity we get 

$$\begin{gathered} \left(a+0.1\right)\left(a-0.1\right)={\left(a\right)}^2-{\left(0.1\right)}^2 \\ =\left(a\times a\right)-\left(0.1\times 0.1\right) \\ =a^2-0.01 \end{gathered}$$

Hence the value of$\left(a+0.1\right)\left(a-0.1\right)$is

(v) The given expression is

We shall use the identity

Here  

By applying identity we get 

Hence the value ofis

Page-4.7

Question 2:

Evaluate each of the following using identities:

(i) (399)²

(ii) (0.98)²

(iii) 991 ☓ 1009

(iv) 117 ☓ 83

Answer 2:

In the given problem, we have to evaluate expressions by using identities.

(i) Given

We can write as

We shall use the Identity  

Where ,

By applying in identity we get 

$$\begin{gathered} {\left(400-1\right)}^2={\left(400\right)}^2-2\times 400\times 1+{\left(1\right)}^2 \\ =400\times 400-800+1 \\ =160000-800+1 \\ =159201 \end{gathered}$$

Hence the value of is

(ii) We have been given

We can write as

We shall use the identity

Where,

By applying in identity we get 

Hence the value of is

(iii) The given expression is

We have

So we can express and in the terms of as 

We shall use the identity $\left(x-y\right)\left(x+y\right)=x^2-y^2$

Here 

By applying in identity we get 

Hence the value of is

(iv) The given expression is

We have

So we can express and in the terms of 100 as

We shall use the identity $\left(x-y\right)\left(x+y\right)=x^2-y^2$

Here

By applying in identity we get 

Hence the value of is

Question 3:

Simplify each of the following:

(i) 175 × 175 + 2 × 175 × 25 + 25 × 25

(ii) 322 × 322 − 2 × 322 × 22 + 22 × 22

(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 × × 0.24

(iv) $\frac{7.83 \times 7.83-1.17\times 1.17}{6.66}$

Answer 3:

In the given problem, we have to simplify expressions

(i) Given

Put

Hence the equation becomes,

That is 

Hence the value of is

(ii) We have been given

Put

Hence the equation becomes

That is

Hence the value of is

(iii) Given

Put

Hence the equation becomes

That is 

Hence the value of is

(iv) We have been given

Put

Hence the equation becomes

Hence the value of is

Question 4:

If $x+\frac{1}{x}=11,$ find the value of $x^2+\frac{1}{x^2}$.

Answer 4:

In the given problem, we have to find

Given

On squaring both sides we get,

Hence the value of is .

Question 5:

If $x-\frac{1}{x}=-1$, find the value of $x^2+\frac{1}{x^2}$

Answer 5:

In the given problem, we have to find

Given

On squaring both sides we get,

We shall use the identity

Hence the value of is.

Question 6:

If $x+\frac{1}{x}=\sqrt{5}$, find the value of $x^2+\frac{1}{x^2}$ and $x^4+\frac{1}{x^4}$.

Answer 6:

In the given problem, we have to find and

We have

On squaring both sides we get,

We shall use the identity

Again squaring on both sides we get,

We shall use the identity

Hence the value ofis and is .

Question 7:

If 9x² + 25y² = 181 and xy = −6, find the value of 3x + 5y.

Answer 7:

In the given problem, we have to find

We have been given and

Let us take

We shall use the identity

 

By substituting and we get,

Hence the value of is.

Question 8:

If 2x + 3y = 8 and xy = 2, find the value of 4x² + 9y²

Answer 8:

In the given problem, we have to find

We have been given and

Let us take

On squaring both sides we get, 

We shall use the identity

By substituting we get,

Hence the value of is

Question 9:

If 3x − 7y = 10 and xy = -1, find the value of 9x² + 49y²

Answer 9:

In the given problem, we have to find

We have been given and

Let us take  

On squaring both sides we get,

We shall use the identity

By substituting we get,

Hence the value of is.

Question 10:

Simplify each of the following products:

(i) $\left(\frac{1}{2}a-3b\right) \left(3b+\frac{1}{2}a\right) \left(\frac{1}{4}{a}^2+9b^2\right)$

(ii) ${\left(m+\frac{n}{7}\right)}^3 \left(m-\frac{n}{7}\right)$

Answer 10:

(i) In the given problem, we have to find product of

We have been given

On rearranging we get,

We shall use the identity

By substituting,we get,

We shall use the identity

Hence the value of is

(ii) In the given problem, we have to find product of

We have been given

On rearranging we get 

We shall use the identity

By substituting,, we get ,

Hence the value of is .

Question 11:

If $x^2+\frac{1}{x^2}=66$, find the value of $x-\frac{1}{x}$

Answer 11:

In the given problem, we have to find

Given

Adding and subtracting 2 on left hand side 

Hence the value of is

Question 12:

If $x^2+\frac{1}{x^2}=79$, find the value of $x+\frac{1}{x}$

Answer 12:

In the given problem, we have to find

Given

Adding and subtracting 2 on left hand side, 

Hence the value of is

Question 13:

Simplify each of the following products:
(i) $\left(\frac{x}{2}-\frac{2}{5}\right) \left(\frac{2}{5}-\frac{x}{2}\right) - x^2+2x$

(ii) $\left(x^2+x-2\right) \left(x^2-x+2\right)$

(iii) $\left(x^3-3x^2-x\right) \left(x^2-3x+1\right)$

(iv) $\left(2x^4-4x^2+1\right) (2x^4-4x^2-1)$

Answer 13:

(i) In the given problem, we have to find product of

On rearranging we get 

We shall use the identity

By substituting ,

Hence the value of is

(ii) In the given problem, we have to find product of

On rearranging we get

We shall use the identity

Hence the value of is

(iii) In the given problem, we have to find product of

Taking as common factor

We shall use the identity

Hence the value of is

(iv) In the given problem, we have to find product of

On rearranging we get

We shall use the identity

Hence the value of is.

Question 14:

Prove that a² + b² + c² −ab−bc−ca is always non-negative for all values of a, b, and c.

Answer 14:

In the given problem, we have to prove is always non negative for all that is we have to prove that

Consider,

Hence is always non negative for all

Note: Square of all negative numbers is always positive or non negative.