RD Sharma 2020 solution class 9 chapter 4 Algebraic Identities Exercise 4.2

Exercise 4.2

Page-4.11

Question 1:

Write the following in the expanded form:

(i) (a +2b + c)²

(ii) (2a − 3b − c)²

(iii) (−3x + y + z)²

(iv) (m + 2n − 5p)²

(v) (2 + x − 2y)²

(vi) (a² + b² + c²)²

(vii) (ab + bc + ca)²

(viii) ${\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)}^2$

(ix) ${\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)}^2$

(x) $(x+2y+4z{)}^2$

(xi) (2x − y + z)²

(xii) (−2x + 3y + 2z)²

Answer 1:

In the given problem, we have to find expended form

(i) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is

(ii) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is

(iii) Given

We shall use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here

By applying in identity we get 

${\left(-3x+y+z\right)}^2={\left(-3x\right)}^2+y^2+z^2-2\times 3x\times y+2yz-2\times \left(-3x\right)\times z$

Hence the expended form of is.

(iv) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is

(v) Given

We shall use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here

By applying in identity we get 

Hence the expended form of is.

(vi) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is.

(vii) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form ofis .

(viii) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is

(ix) Given

We shall use the identity

Here

By applying in identity we get 

Hence the expended form of is

(x) Given

We shall use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here

By applying in identity we get 

Hence the expended form of is

(xi) Given

We shall use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here

By applying in identity we get 

Hence the expended form of is

(xii) Given

We shall use the identity ${\left(a+b+c\right)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here

By applying in identity we get 

Hence the expended form of is.

Page-4.12

Question 2:

If a + b + c = 0 and a² + b² + c² = 16, find the value of ab + bc + ca.

Answer 2:

In the given problem, we have to find value of

Given and

Squaring the equation, we get

Now putting the value of in above equation we get,

Taking 2 as common factor we get 

Hence the value of is .

Question 3:

If a² + b² + c² = 16 and ab + bc + ca = 10, find the value of a + b + c.

Answer 3:

In the given problem, we have to find value of

Given

Multiply equation with 2 on both sides we get,

Now adding both equation and we get 

We shall use the identity

Hence the value of is .

Question 4:

If a + b + c = 9 and ab + bc + ca = 23, find the value of a² + b² + c².

Answer 4:

In the given problem, we have to find value of

Given

Squaring both sides of we get,

Substituting in above equation we get,

Hence the value of is.

Question 5:

Find the value of 4x² + y² + 25x² + 4xy − 10yz − 20zx when x= 4, y = 3 and z = 2.

Answer 5:

In the given problem, we have to find value of

Given

We have

This equation can also be written as

Using the identity

${\left(x+y-z\right)}^2=x^2+y^2+z^2+2xy-2yz-2xz$

Hence the value of is .

Question 6:

Simplify:

(i) (a +b + c)² + (a − b + c)²

(ii) (a +b + c)² − (a − b + c)2

(iii) (a +b + c)² + (a − b + c)² + (a +b − c)²

(iv) (2x + p − c)² − (2x − p + c)²

(v) (x² + y² − z²) − (x² − y² + z²)²

Answer 6:

In the given problem, we have to simplify the expressions

(i) Given

By using identity

Hence the equation becomes 

Taking 2 as common factor we get 

Hence the simplified value of is

(ii) Given

By using identity

Hence the equation becomes

 

Taking 4 as common factor we get 

Hence the simplified value of is.

(iii) Given

By using identity , we have

Taking 3 as a common factor we get 

Hence the value ofis 

.

(iv) Given

By using identity , we get

By cancelling the opposite terms, we get 

Taking as common a factor we get,

Hence the value of is

(v) We have (x² + y² − z²) − (x² − y² + z²)²

Using formula, we get

(x² + y² − z²) − (x² − y² + z²)²

 

By canceling the opposite terms, we get 

Taking as common factor we get 

Hence the value of is.

Question 7:

Simplify each of the following expressions:

(i) $(x+y+z{)}^2+{\left(x+\frac{y}{2}+\frac{z}{3}\right)}^2-{\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)}^2$

(ii) (x + y− 2z)² − x² − y² − 3z² + 4xy

(iii) (x²− x + 1)² − (x² + x + 1)²

Answer 7:

In the given problem, we have to simplify the value of each expression 

(i) Given

We shall use the identity for each bracket

$-\left\{{\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{3}\right)}^2+{\left(\frac{z}{4}\right)}^2+2\left(\frac{x}{2}\right)\left(\frac{y}{3}\right)+2\left(\frac{y}{3}\right)\left(\frac{z}{4}\right)+2\left(\frac{x}{2}\right)\left(\frac{z}{4}\right)\right\}$

By arranging the like terms we get

Now adding or subtracting like terms,

Hence the value of is

(ii) Given

We shall use the identity ${\left(x+y-z\right)}^2=x^2+y^2+z^2+2xy-2yz-2zx$ for expanding the brackets

Now arranging liked terms we get,

Hence the value of is

(iii) Given

We shall use the identity for each brackets

$$\begin{gathered} {\left(x^2-x+1\right)}^2-{\left(x^2+x+1\right)}^2=\left[{\left(x^2\right)}^2+{\left(-x\right)}^2+1^2-2x^3-2x+2x^2\right] \\ -\left[{\left(x^2\right)}^2+{\left(x\right)}^2+1^2+2x^3+2x+2x^2\right] \end{gathered}$$

Canceling the opposite term and simplifies

Hence the value of is .