NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Exploring Algebraic Identities Class 9 Questions and Answers PDF Free Download

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NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities: Exercise Questions

Below, you will find the NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities question answers explained step by step.

Think and Reflect (Page 71)

Question 1. What can you say about $a$ and $b$ if $(a+b)^2<a^2+b^2$ ?

$\textbf{Answer:}$

$(a+b)^2<a^2+b^2$
$⇒a^2+2 a b+b^2<a^2+b^2$
$⇒2ab<0$
$⇒ab<0$
So, $a$ and $b$ have opposite signs (one positive and one negative).

Question 2. What can you say about $a$ and $b$ if $(a+b)^2>a^2+b^2$ ?

$\textbf{Answer:}$

$(a+b)^2>a^2+b^2$
$⇒a^2+2 a b+b^2>a^2+b^2$
$⇒2ab>0$
$⇒ab>0$
So, $a$ and $b$ have the same signs (both positive and both negative).

Question 3. When will $(a+b)^2$ be equal to $a^2+b^2$ ?

$\textbf{Answer:}$

If $(a+b)^2=a^2+b^2$
$⇒2ab=0$
$⇒ab=0$
So, either $a=0$ or $b=0$ (or both).

Did you observe that $(a+b)^2$ and $a^2+b^2$ are both positive? What term will decide which is larger? Use the expansion of $(a+b)^2$ to decide.

The deciding term is: $2ab$

So,
$2 a b>0 \Rightarrow(a+b)^2$ is larger
$2 a b<0 \Rightarrow a^2+b^2$ is larger
$2 a b=0 \Rightarrow$ both are equal

Question 1. Using the identity $(a+b)^2=a^2+2 a b+b^2$, expand the following:
(i) $(7 x+4 y)^2$
(ii) $\left(\frac{7}{5} x+\frac{3}{2} y\right)^2$
(iii) $(2.5 p+1.5 q)^2$
(iv) $\left(\frac{3}{4} s+8 t\right)^2$
(v) $\left(x+\frac{1}{2 y}\right)^2$
(vi) $\left(\frac{1}{x}+\frac{1}{y}\right)^2$

$\textbf{Answer:}$

(i) $(7 x+4 y)^2$
$=(7 x)^2+2(7 x)(4 y)+(4 y)^2$
$=49x^2+56xy+16y^2$

(ii) $\left(\frac{7}{5} x+\frac{3}{2} y\right)^2$
$=\left(\frac{7}{5} x\right)^2+2\left(\frac{7}{5} x\right)\left(\frac{3}{2} y\right)+\left(\frac{3}{2} y\right)^2$
$=\frac{49}{25} x^2+\frac{21}{5} x y+\frac{9}{4} y^2$

(iii) $(2.5 p+1.5 q)^2$
$=(2.5 p)^2+2(2.5 p)(1.5 q)+(1.5 q)^2$
$=6.25p^2+7.5 pq+2.25q^2$

(iv) $\left(\frac{3}{4} s+8 t\right)^2$
$=\left(\frac{3}{4} s\right)^2+2\left(\frac{3}{4} s\right)(8 t)+(8 t)^2$
$=\frac{9}{16} s^2+12 s t+64 t^2$

(v) $\left(x+\frac{1}{2 y}\right)^2$
$=x^2+2(x)\left(\frac{1}{2 y}\right)+\left(\frac{1}{2 y}\right)^2$
$=x^2+\frac{x}{y}+\frac{1}{4 y^2}$

(vi) $\left(\frac{1}{x}+\frac{1}{y}\right)^2$
$=\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\left(\frac{1}{y}\right)^2$
$=\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2}$

Question 2. Using the same identity, find the values of the following:
(i) $(64)^2$
(ii) $(105)^2$
(iii) $(205)^2$

$\textbf{Answer:}$

(i) $(64)^2$
$=(60+4)^2$
$=60^2+2(60)(4)+4^2$
$=3600+480+16$
$=4096$

(ii) $(105)^2$
$=(100+5)^2$
$=100^2+2\times100\times5+5^2$
$=10000+1000+25$
$=11025$

(iii) $(205)^2$
$=(200+5)^2$
$=200^2+2\times200\times5+5^2$
$=40000+2000+25$
$=42025$

Think and Reflect (Page 73)

Question: What if we replace $b$ by $-b$ in $(a+b)^2=a^2+2 a b+b^2$ ?

$\textbf{Answer:}$

If we replace $b$ by $-b$ in $(a+b)^2=a^2+2 a b+b^2$, we get,
$(a+(-b))^2=a^2+2 a(-b)+(-b)^2$
$⇒(a-b)^2=a^2-2ab+b^2$

$\therefore$ Only the middle term changes sign.

Question 1. Factor completely:
(i) $9 x^2+24 x y+16 y^2$
(ii) $4 s^2+20 s t+25 t^2$
(iii) $49 x^2+28 x y+4 y^2$
(iv) $64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2$
*(v) $ 3 a^2+4 a b+\frac{4}{3} b^2$
*(vi) $ \frac{9}{5} s^2+6 s v+5 v^2$
(Hint: 2 was taken out as a common factor in Example 7. Is it possible to do something similar in Exercises (v) and (vi) above?)

$\textbf{Answer:}$

(i) $9 x^2+24 x y+16 y^2$
$=(3 x)^2+2(3 x)(4 y)+(4 y)^2$
$=(3x+4y)^2$

(ii) $4 s^2+20 s t+25 t^2$
$=(2 s)^2+2(2 s)(5 t)+(5 t)^2$
$=(2s+5t)^2$

(iii) $49 x^2+28 x y+4 y^2$
$=(7 x)^2+2(7 x)(2 y)+(2 y)^2$
$=(7x+2y)^2$

(iv) $64 p^2+\frac{32}{3} p q+\frac{4}{9} q^2$
$=(8 p)^2+2(8 p)\left(\frac{2}{3} q\right)+\left(\frac{2}{3} q\right)^2$
$=(8p+\frac23q)^2$

*(v) $ 3 a^2+4 a b+\frac{4}{3} b^2$
$=3\left(a^2+\frac{4}{3} a b+\frac{4}{9} b^2\right)$
$=3(a^2+\frac{4}{3} a b+\frac{4}{9} b^2)$
$=3\left(a+\frac{2}{3} b\right)^2$

*(vi) $ \frac{9}{5} s^2+6 s v+5 v^2$
$=\frac{1}{5}\left(9 s^2+30 s v+25 v^2\right)$
$=\frac15(9 s^2+30 s v+25 v^2)$
$=\frac15(3 s+5 v)^2$

Question 2. Find the values of the following using the identity $(a-b)^2=a^2-2 a b+b^2$.
(i) $(79)^2$
(ii) $(193)^2$
(iii) $(299)^2$

$\textbf{Answer:}$

(i) $(79)^2$
$=(80-1)^2$
$=80^2-2(80)(1)+1^2$
$=6400-160+1$
$=6241$

(ii) $(193)^2$
$=(200-7)^2$
$=200^2-2\times200\times7+7^2$
$=40000-2800+49$
$=37249$

(iii) $(299)^2$
$=(300-1)^2$
$=300^2-2\times300\times1+1^2$
$=90000-600+1$
$=89401$

Question 1. Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.
(i) $117^2$
(ii) $78^2$
(iii) $198^2$
(iv) $214^2$
(v) $1104^2$
(vi) $1120^2$

$\textbf{Answer:}$
We will use the following identities to make the squaring process easier.
$\begin{aligned} & (a+b)^2=a^2+2 a b+b^2 \\ & (a-b)^2=a^2-2 a b+b^2\end{aligned}$

(i) $117^2$
$=(100+17)^2$
$=100^2+2\times100\times17+17^2$
$=10000+3400+289$
$=13689$

(ii) $78^2$
$=(80-2)^2$
$=80^2-2\times80\times2+2^2$
$=6400-320+4$
$=6084$

(iii) $198^2$
$=(200-2)^2$
$=200^2-2\times200\times2+2^2$
$=40000-800+4$
$=39204$

(iv) $214^2$
$=(200+14)^2$
$=200^2+2\times200\times14+14^2$
$=40000+5600+196$
$=45796$

(v) $1104^2$
$=(1100+4)^2$
$=1100^2+2\times1100\times4+4^2$
$=1210000+8800+16$
$=1218816$

(vi) $1120^2$
$=(1100+20)^2$
$=1100^2+2\times1100\times20+20^2$
$=1210000+44000+400$
$=1254400$

Question 2. Factor using suitable identities:
(i) $16 y^2-24 y+9$
(ii) $\frac{9}{4} s^2+6 s t+4 t^2$
(iii) $\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2$
(iv) $\frac{p^2}{16}-2+\frac{16}{p^2}$
(v) $9 a^2+4 b^2+c^2-12 a b+6 a c-4 b c$

$\textbf{Answer:}$

(i) $16 y^2-24 y+9$
$=(4 y)^2-2(4 y)(3)+3^2$
$=(4y-3)^2$

(ii) $\frac{9}{4} s^2+6 s t+4 t^2$
$=\left(\frac{3}{2} s\right)^2+2\left(\frac{3}{2} s\right)(2 t)+(2 t)^2$
$=(\frac32s+2t)^2$

(iii) $\frac{m^2}{9}+\frac{m k}{3}+\frac{k^2}{4}+3 n k+2 m n+9 n^2$
We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
We will convert the equation into the form of the above identity.
$\frac{m^2}{9}+\frac{k^2}{4}+9 n^2+\frac{m k}{3}+3 n k+2 m n$
$=(\frac m3)^2+(\frac k2)^2+(3n)^2+2\left(\frac{m}{3}\right)\left(\frac{k}{2}\right)+2\left(\frac{k}{2}\right)(3 n)+2(3 n)\left(\frac{m}{3}\right)$
$=\left(\frac{m}{3}+\frac{k}{2}+3 n\right)^2$

(iv) $\frac{p^2}{16}-2+\frac{16}{p^2}$
$=\left(\frac{p}{4}\right)^2-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)+\left(\frac{4}{p}\right)^2$
$=(\frac p4-\frac 4p)^2$

(v) $9 a^2+4 b^2+c^2-12 a b+6 a c-4 b c$
We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
We will convert the equation into the form of the above identity.
$9 a^2+4 b^2+c^2-12 a b+6 a c-4 b c$
$=(3 a)^2+(-2 b)^2+c^2+2(3 a)(-2 b)+2(3 a)(c)+2(-2 b)(c)$
$=(3a-2b+c)^2$

Question 3. Expand the following using the identity
$(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$ :
(i) $(p+3 q+7 r)^2$
(ii) $(3 x-2 y+4 z)^2$

$\textbf{Answer:}$

We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$

(i) $(p+3 q+7 r)^2$
$=p^2+(3 q)^2+(7 r)^2+2(p)(3 q)+2(3 q)(7 r)+2(7 r)(p)$
$=p^2+9 q^2+49 r^2+6 p q+42 q r+14 p r$

(ii) $(3 x-2 y+4 z)^2$
$=(3 x)^2+(-2 y)^2+(4 z)^2+2(3 x)(-2 y)+2(-2 y)(4 z)+2(4 z)(3 x)$
$=9 x^2+4 y^2+16 z^2-12 x y-16 y z+24 x z$

Question 4. Is this an identity?

$(a+b-c)^2+(a-b+c)^2+(a-b-c)^2=2 a^2+2 b^2+2 c^2$

$\textbf{Answer:}$

LHS
$=(a+b-c)^2+(a-b+c)^2+(a-b-c)^2$
$=a^2+b^2+c^2a^2+b^2+c^2+2 a b-2 b c-2 a c-2ab-2bc$
$+2ac+a^2+b^2+c^2-2ab+2bc-2ac$
$=3 a^2+3 b^2+3 c^2-2 a b-2 b c-2 a c$
$\neq$ RHS

So, it is not an identity.

Think and Reflect (Page 78)

Question 1. Try to evaluate the following using a suitable identity:

(i) $35^2$

(ii) $65^2$

(iii) $85^2$

(iv) $105^2$

Do you observe any interesting pattern?

$\textbf{Answer:}$
(i)
$
\begin{aligned}
& 35^2=(30+5)^2 \\
& \Rightarrow 35^2=30^2+2(30)(5)+5^2 \\
& \Rightarrow 35^2=900+300+25 \\
& \Rightarrow 35^2=1225
\end{aligned}
$
Hence, the correct answer is 1225.

(ii)
$
\begin{aligned}
& 65^2=(60+5)^2 \\
& \Rightarrow 65^2=60^2+2(60)(5)+5^2 \\
& \Rightarrow 65^2=3600+600+25 \\
& \Rightarrow 65^2=4225
\end{aligned}
$
Hence, the correct answer is 4225.
(iii)
$
\begin{aligned}
& 85^2=(80+5)^2 \\
& \Rightarrow 85^2=80^2+2(80)(5)+5^2 \\
& \Rightarrow 85^2=6400+800+25 \\
& \Rightarrow 85^2=7225
\end{aligned}
$
Hence, the correct answer is 7225.
(iv)
$
\begin{aligned}
& 105^2=(100+5)^2 \\
& \Rightarrow 105^2=100^2+2(100)(5)+5^2 \\
& \Rightarrow 105^2=10000+1000+25 \\
& \Rightarrow 105^2=11025
\end{aligned}
$
Hence, the correct answer is 11025.

There is a pattern for squaring any number ending in 5.
For any number of the form $10 n+5$ (where $n$ is the part of the number before the 5):

• The last two digits of the result are always 25.
• The preceding digits are calculated by multiplying $n$ with its successor, ( $n+1$ ).

$(10 n+5)^2=n(n+1) \times 100+25$

Question 2. Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.

$\textbf{Answer:}$

Top Row Shapes:

• A large composite shape with an outer boundary side length of ( $a+b+c$ ), containing a green square of side $2 b$, a white square of side $2 c$, and two red regions.
• A yellow square with side length $(a+b-c) \Rightarrow$ Area $=(a+b-c)^2$
• A blue square with side length $(a-b+c) \Rightarrow \operatorname{Area}=(a-b+c)^2$
• A purple square with side length $(a-b-c) \Rightarrow$ Area $=(a-b-c)^2$

Bottom Row Shapes:

• A large composite square with a side length of 2a, perfectly reassembled using the two red regions, the yellow square, the blue square, and the purple square ⇒ Area $=(2 a)^2$
• The green square with side length $2 b \Rightarrow$ Area $=(2 b)^2$
• The white square with side length $2 c \Rightarrow \operatorname{Area}=(2 c)^2$

Question 1. Fill in the blanks to complete the following identities:
(i) $s^2-11 s+24=($ $\_\_\_\_$ ) ( $\_\_\_\_$ )
(ii) $($ $\_\_\_\_$ $(x+1)=\left(3 x^2-4 x-7\right)$
(iii) $10 x^2-11 x-6=(2 x-$ $\_\_\_\_$ ) ( $\_\_\_\_$ + 2)
(iv) $6 x^2+7 x+2=($ $\_\_\_\_$ ) ( $\_\_\_\_$ )

$\textbf{Answer:}$

(i)
$s^2-11 s+24$
$=s^2-3s-8s+24$
$=(s-3)(s-8)$
So, $s^2-11 s+24=(s-3)(s-8)$

(ii)
RHS
$=3 x^2-4 x-7$
$=3 x^2+3 x-7 x-7$
$=(3 x-7)(x+1)$
So, blank space is $(3x-7)$.

(iii)
$10 x^2-15 x+4 x-6 $
$=5 x(2 x-3)+2(2 x-3) $
$=(2 x-3)(5 x+2)$
So, $10 x^2-11 x-6=(2 x-3)(5 x+2)$
$\therefore$ Missing numbers are 3 and $5x$

(iv)
$6 x^2+3 x+4 x+2 $
$=3 x(2 x+1)+2(2 x+1) $
$ =(3 x+2)(2 x+1)$

Question 2. Select and use the identity that will help you to find the following products without multiplying directly:
(i) $(41)^2$
(ii) $(27)^2$
(iii) $(23 \times 17)$
(iv) $(135)^2$
(v) $(97)^2$
(vi) $(18 \times 29)$
(vii) $(34 \times 43)$
(viii) $(205)^2$

$\textbf{Answer:}$

(i) $(41)^2$
$=(40+1)^2$
$=40^2+2\times40\times1+1^2$
$=1600+80+1$
$=1681$

(ii) $(27)^2$
$=(30-3)^2$
$=30^2-2\times30\times3+3^2$
$=900-180+9$
$=729$

(iii) $(23 \times 17)$
We know that $(a+b)(a-b)=a^2-b^2$
$(23 \times 17)=(20+3)(20-3)=20^2-3^2=400-9=391$

(iv) $(135)^2$
$=(100+35)^2$
$=100^2+2\times100\times35+35^2$
$=10000+7000+1225$
$=18225$

(v) $(97)^2$
$=(100-3)^2$
$=100^2-2\times100\times3+3^2$
$=10000-600+9$
$=9409$

(vi) $(18 \times 29)$
$=18\times(30-1)$
$=540-18$
$=522$

(vii) $(34 \times 43)$
We know that $(a+b)(a-b)=a^2-b^2$
$=(38.5-4.5)(38.5+4.5)$
$=38.5^2-4.5^2$
$=1482.25-20.25$
$=1462$

(viii) $(205)^2$
$=(200+5)^2$
$=200^2+2\times200\times5+5^2$
$=40000+2000+25$
$=42025$

Question 3. Factor the following:
(i) $9 a^2+b^2+4 c^2-6 a b+12 a c-4 b c$
(ii) $16 s^2+25 t^2-40 s t$
(iii) $r^2-r-42$
(iv) $49 g^2+14 g h+h^2$
(v) $64 u^2+121 v^2+4 w^2-176 u v-32 u w+44 v w$

$\textbf{Answer:}$

(i) $9 a^2+b^2+4 c^2-6 a b+12 a c-4 b c$
We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$=(3a)^2+b^2+(2c)^2+ 2(3 a)(-b) + 2(-b)(2 c) +2(2 c)(3 a)$
$=(3 a-b+2 c)^2$

(ii) $16 s^2+25 t^2-40 s t$
$=(4 s)^2+(5 t)^2-2(4 s)(5 t)$
$ =(4 s-5 t)^2$

(iii) $r^2-r-42$
$r^2-7 r+6 r-42 $
$ =r(r-7)+6(r-7) $
$ =(r-7)(r+6)$

(iv) $49 g^2+14 g h+h^2$
$=(7 g)^2+2(7 g)(h)+h^2$
$=(7 g+h)^2$

(v) $64 u^2+121 v^2+4 w^2-176 u v-32 u w+44 v w$
We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$=(8 u)^2+(-11 v)^2+(-2 w)^2+2(8 u)(-11 v)+2(-11 v)(-2 w)+2(-2 w)(8 u)$
$=(8 u-11 v-2 w)^2$

Think and Reflect (Page 82)

Question: James and Reshma were talking about algebraic identities they learnt in school.

James: $(a-b)^2(a+b)=\left(a^2-2 a b+b^2\right)(a+b)$
Reshma: I have a different idea. $(a-b)^2(a+b)=(a-b)[(a-b)(a+b)]$

$=(a-b)\left(a^2-b^2\right)$
I will find this product to get the answer.
According to you, who is correct and why?
Try to combine more such identities and find new results.

$\textbf{Answer:}$

James expands the squared term first:
$
\begin{aligned}
& (a-b)^2(a+b)=\left(a^2-2 a b+b^2\right)(a+b) \\
& \Rightarrow(a-b)^2(a+b)=a\left(a^2-2 a b+b^2\right)+b\left(a^2-2 a b+b^2\right) \\
& \Rightarrow(a-b)^2(a+b)=a^3-2 a^2 b+a b^2+a^2 b-2 a b^2+b^3 \\
& \Rightarrow(a-b)^2(a+b)=a^3-a^2 b-a b^2+b^3
\end{aligned}
$
Hence, James's method is correct.
Reshma regrouped the factors using the associative property to create a difference of squares:
$
\begin{aligned}
& (a-b)^2(a+b)=(a-b)[(a-b)(a+b)] \\
& \Rightarrow(a-b)^2(a+b)=(a-b)\left(a^2-b^2\right) \\
& \Rightarrow(a-b)^2(a+b)=a\left(a^2-b^2\right)-b\left(a^2-b^2\right) \\
& \Rightarrow(a-b)^2(a+b)=a^3-a b^2-a^2 b+b^3 \\
& \Rightarrow(a-b)^2(a+b)=a^3-a^2 b-a b^2+b^3
\end{aligned}
$
Hence, Reshma's method is also correct and yields the exact same final polynomial.

Let's add the expansions of $(a+b)^2$ and $(a-b)^2$ :

$
\begin{aligned}
& (a+b)^2+(a-b)^2=\left(a^2+2 a b+b^2\right)+\left(a^2-2 a b+b^2\right) \\
& \Rightarrow(a+b)^2+(a-b)^2=2 a^2+2 b^2 \\
& \Rightarrow(a+b)^2+(a-b)^2=2\left(a^2+b^2\right)
\end{aligned}
$
Hence, we get a new identity for the sum of squares.

Let's evaluate the product of a sum and difference of cubes, $\left(a^3-b^3\right)\left(a^3+b^3\right)$, using the difference of squares identity $x^2-y^2=(x-y)(x+y)$ :
Let $x=a^3$ and $y=b^3$.
$
\begin{aligned}
& \left(a^3-b^3\right)\left(a^3+b^3\right)=\left(a^3\right)^2-\left(b^3\right)^2 \\
& \Rightarrow\left(a^3-b^3\right)\left(a^3+b^3\right)=a^6-b^6
\end{aligned}
$
Hence, the product simplifies to a difference of sixth powers.

Question 1. Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

(i) $\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}$

(ii) $\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}$

(iii) $\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$

(iv) $\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}$

(v) $\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}$

(vi) $\frac{p^4-16}{p^2-4 p+4}$

$\textbf{Answer:}$

(i) $\frac{3 p^2-3 p q-18 q^2}{p^2+3 p q-10 q^2}$
$=\frac{3( p^2- p q-6 q^2)}{p^2+3 p q-10 q^2}$
$=\frac{3( p^2- 3p q+2pq-6 q^2)}{p^2+5 p q-2pq-10 q^2}$
$=\frac{3(p-3 q)(p+2 q)}{(p+5q)(p-2q)}$

(ii) $\frac{n^3-3 n^2 m+3 n m^2-m^3}{5 m^2-10 m n+5 n^2}$
$=\frac{(n-m)^3}{5\left(m^2-2 m n+n^2\right)} $
$ =\frac{(n-m)^3}{5(m-n)^2} $
$ =\frac{(n-m)^3}{5(n-m)^2} $
$ =\frac{n-m}{5}$

(iii) $\frac{w^3-v^3+x^3+3 w v x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$
We know that $a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=\frac{w^3+(-v)^3+x^3-3 w (-v) x}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$
$=\frac{(w+x-v)\left(w^2+x^2+v^2-w x+v x+w v\right)}{w^2+v^2+x^2-2 w v-2 v x+2 w x}$
Also, $w^2+v^2+x^2-2 w v-2 v x+2 w x=(w-v+x)^2 $
$=\frac{(w-v+x)\left(w^2+x^2+v^2-w x+v x+w v\right)}{(w-v+x)^2}$
$=\frac{w^2+x^2+v^2-w x+v x+w v}{w-v+x}$

(iv) $\frac{4 y^2-20 y z+25 z^2}{\left(25 z^2-4 y^2\right)}$
$=\frac{(2 y)^2-2(2 y)(5 z)+(5 z)^2}{(5 z)^2-(2 y)^2} $
$ =\frac{(2 y-5 z)^2}{(5 z-2 y)(5 z+2 y)} $
$ =\frac{(5 z-2 y)^2}{(5 z-2 y)(5 z+2 y)} $
$=\frac{5 z-2 y}{5 z+2 y}$

(v) $\frac{\left(x^2+x-6\right)\left(x^2-7 x+12\right)}{\left(x^2-6 x+8\right)\left(x^2-9\right)}$
$=\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}$
$=1$

(vi) $\frac{p^4-16}{p^2-4 p+4}$
$ =\frac{\left(p^2\right)^2-4^2}{p^2-4 p+4} $
$= \frac{\left(p^2-4\right)\left(p^2+4\right)}{(p-2)^2} $
$= \frac{(p-2)(p+2)\left(p^2+4\right)}{(p-2)^2} $
$= \frac{(p+2)\left(p^2+4\right)}{p-2}$

Question 1. Use suitable identities to find the following products:
(i) $(-3 x+4)^2$
(ii) $(2 s+7)(2 s-7)$
(iii) $\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)$
(iv) $(2 n+7)(2 n-7)$
(v) $(s-2 t)\left(s^2+2 s t+4 t^2\right)$
(vi) $\left(\frac{1}{2 r}-4 r\right)^2$
(vii) $(-3 m+4 k-l)^2$
(viii) $\left(x-\frac{1}{3} y\right)^3$
(ix) $\left(\frac{7}{2} k-\frac{2}{3} m\right)^3$

$\textbf{Answer:}$

(i) $(-3 x+4)^2$
$=(-3 x)^2+2(-3 x)(4)+4^2 $
$=9 x^2-24 x+16$

(ii) $(2 s+7)(2 s-7)$
$=(2s)^2-7^2$
$=4s^2-49$

(iii) $\left(p^2+\frac{1}{2}\right)\left(p^2-\frac{1}{2}\right)$
$=(p^2)^2-(\frac12)^2$
$=p^4-\frac14$

(iv) $(2 n+7)(2 n-7)$
$=(2n)^2-7^2$
$=4n^2-49$

(v) $(s-2 t)\left(s^2+2 s t+4 t^2\right)$
We know, $(a-b)\left(a^2+a b+b^2\right)=a^3-b^3$
$=s^3-(2 t)^3 $
$ =s^3-8 t^3$

(vi) $\left(\frac{1}{2 r}-4 r\right)^2$
$=\left(\frac{1}{2 r}\right)^2-2\left(\frac{1}{2 r}\right)(4 r)+(4 r)^2$
$ =\frac{1}{4 r^2}-4+16 r^2$

(vii) $(-3 m+4 k-l)^2$
We know that $(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a$
$=(-3 m)^2+(4 k)^2+(-l)^2+2(-3 m)(4 k)+2(4 k)(-l)+2(-l)(-3 m) $
$ =9 m^2+16 k^2+l^2-24 m k-8 k l+6 m l$

(viii) $\left(x-\frac{1}{3} y\right)^3$
We know that $(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3$
$=x^3-3 x^2\left(\frac{1}{3} y\right)+3 x\left(\frac{1}{3} y\right)^2-\left(\frac{1}{3} y\right)^3 $
$ =x^3-x^2 y+\frac{1}{3} x y^2-\frac{1}{27} y^3$

(ix) $\left(\frac{7}{2} k-\frac{2}{3} m\right)^3$
We know that $(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3$
$=\left(\frac{7}{2} k\right)^3-3\left(\frac{7}{2} k\right)^2\left(\frac{2}{3} m\right)+3\left(\frac{7}{2} k\right)\left(\frac{2}{3} m\right)^2-\left(\frac{2}{3} m\right)^3 $
$=\frac{343}{8} k^3-\frac{49}{2} k^2 m+\frac{14}{3} k m^2-\frac{8}{27} m^3$

Question 2. Find the values using suitable identities:
(i) $17 \times 21$
(ii) $104 \times 96$
(iii) $24 \times 16$
(iv) $147^3$
(v) $199^3$
(vi) $127^3$
(vii) $(-107)^3$
(viii) $(-299)^3$

$\textbf{Answer:}$

(i) $17 \times 21$
$=(19-2)(19+2)$
$=19^2-3^2$
$=361-9$
$=352$

(ii) $104 \times 96$
$=(100+4)(100-4)$
$=100^2-4^2$
$=10000-16$
$=9984$

(iii) $24 \times 16$
$=(20+4)(20-4)$
$=20^2-4^2$
$=400-16$
$=384$

(iv) $147^3$
$=(150-3)^3$
$=150^3-3(150)^2(3)+3(150)(3)^2-3^3$
$ =3375000-202500+4050-27 $
$ =3176523$

(v) $199^3$
$=(200-1)^3$
$=200^3-3\times200^2\times1+3\times200\times1^2-1^3$
$=8000000-120000+600-1$
$=7880599$

(vi) $127^3$
$=(100+27)^3$
$=100^3+3\times100^2\times3+3\times100\times3^2+3^3$
$=1000000+810000+218700+19683$
$=2048383$

(vii) $(-107)^3$
$=-(100+7)^3$
$=-(1000000+210000+14700+343) $
$ =-1225043$

(viii) $(-299)^3$
$=(300-1)^3$
$=-(27000000-270000+900-1) $
​​​​​​​$ =-26730901$

Question 3. Factor the following algebraic expressions:
(i) $4 y^2+1+\frac{1}{16 y^2}$
(ii) $9 m^2-\frac{1}{25 n^2}$
(iii) $ 27 b^3-\frac{1}{64 b^3}$
(iv) $x^2+\frac{5 x}{6}+\frac{1}{6}$
(v) $27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}$
(vi) $64 y^3+\frac{1}{125} z^3$
(vii) $p^3+27 q^3+r^3-9 p q r$
(viii) $9 m^2-12 m+4$
(ix) $9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z$
(x) $4 x^2+9 y^2+36 z^2+12 x z+36 y z+24 x y$
(xi) $27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}$

$\textbf{Answer:}$

(i) $4 y^2+1+\frac{1}{16 y^2}$
$=(2 y)^2+2(2 y)\left(\frac{1}{4 y}\right)+\left(\frac{1}{4 y}\right)^2 $
$=\left(2 y+\frac{1}{4 y}\right)^2$

(ii) $9 m^2-\frac{1}{25 n^2}$
$=(3 m)^2-\left(\frac{1}{5 n}\right)^2 $
$=\left(3 m-\frac{1}{5 n}\right)\left(3 m+\frac{1}{5 n}\right)$

(iii) $27 b^3-\frac{1}{64 b^3}$
$=(3 b)^3-\left(\frac{1}{4 b}\right)^3$
$=\left(3 b-\frac{1}{4 b}\right)\left(9 b^2+\frac{3}{4}+\frac{1}{16 b^2}\right)$

(iv) $x^2+\frac{5 x}{6}+\frac{1}{6}$
$=x^2+\frac{2 x}{3}+\frac{x}{6}+\frac{1}{6}$
$=x\left(x+\frac{2}{3}\right)+\frac{1}{6}(x+1)$
$ =\left(x+\frac{1}{2}\right)\left(x+\frac{1}{3}\right)$

(v) $27 u^3-\frac{1}{125}-\frac{27 u^2}{5}+\frac{9 u}{25}$
$=(3 u)^3-\left(\frac{1}{5}\right)^3-3(3 u)^2\left(\frac{1}{5}\right)+3(3 u)\left(\frac{1}{5}\right)^2 $
$=\left(3 u-\frac{1}{5}\right)^3$

(vi) $64 y^3+\frac{1}{125} z^3$
$=(4 y)^3+\left(\frac{z}{5}\right)^3 $
​​​​​​​$=\left(4 y+\frac{z}{5}\right)\left(16 y^2-\frac{4 y z}{5}+\frac{z^2}{25}\right)$

(vii) $p^3+27 q^3+r^3-9 p q r$
$=p^3+(3 q)^3+r^3-3(p)(3 q)(r) $
​​​​​​​$=(p+3 q+r)\left(p^2+9 q^2+r^2-3 p q-3 q r-p r\right)$

(viii) $9 m^2-12 m+4$

$=(3 m)^2-2(3 m)(2)+2^2 $

​​​​​​​$=(3 m-2)^2$

(ix) $9 x^3-\frac{8}{3} y^3+\frac{z^3}{3}+6 x y z$
$=\frac13\left[9 x^3-8 y^3+z^3+18 x y z\right] $
$ =\frac13\left[(3 x)^3+(-2 y)^3+z^3-3(3 x)(-2 y) z\right] $
$=\frac13(3 x-2 y+z)\left[(3 x)^2+(-2 y)^2+z^2-(3 x)(-2 y)-(-2 y)(z)-(z)(3 x)\right] $ Using $\left.a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]$, we get,
$ =\frac13(3 x-2 y+z)\left[9 x^2+4 y^2+z^2+6 x y+2 y z-3 z x\right]$

(x) $4 x^2+9 y^2+36 z^2+12 x z+36 y z+24 x y$
$=(2 x)^2+(3 y)^2+(6 z)^2+2(2 x)(3 y)+2(3 y)(6 z)+2(6 z)(2 x) $
$=(2 x+3 y+6 z)^2$

(xi) $27 u^3-\frac{1}{216}-\frac{9 u^2}{2}+\frac{u}{4}$
$=(3 u)^3-\left(\frac{1}{6}\right)^3-3(3 u)^2\left(\frac{1}{6}\right)+3(3 u)\left(\frac{1}{6}\right)^2 $
$ =\left(3 u-\frac{1}{6}\right)^3$

Question 4. Simplify the following:
(i) $\frac{4 x^2+4 x+1}{4 x^2-1}$
(ii) $\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}$
(iii) $\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}$

Note: Assume that the denominators are not equal to 0.

$\textbf{Answer:}$

(i) $\frac{4 x^2+4 x+1}{4 x^2-1}$
$=\frac{(2 x+1)^2}{(2 x-1)(2 x+1)} $
$ =\frac{2 x+1}{2 x-1}$

(ii) $\frac{9\left(3 a^3-24 b^3\right)}{9 a^2-36 b^2}$
$=\frac{27\left(a^3-8 b^3\right)}{9\left(a^2-4 b^2\right)}$
$=\frac{27(a-2 b)\left(a^2+2 a b+4 b^2\right)}{9(a-2 b)(a+2 b)} $
​​​​​​​$=\frac{3\left(a^2+2 a b+4 b^2\right)}{a+2 b}$

(iii) $\frac{s^3+125 t^3}{s^2-2 s t-35 t^2}$
$=\frac{(s+5 t)\left(s^2-5 s t+25 t^2\right)}{(s-7 t)(s+5 t)}$
$=\frac{s^2-5 s t+25 t^2}{s-7 t}$

Question 5. Find possible expressions for the length and breadth of each of the following rectangles whose areas are given by the following expressions in square units.
(i) $25 a^2-30 a b+9 b^2$
(ii) $36 s^2-49 t^2$

$\textbf{Answer:}$

(i) $25 a^2-30 a b+9 b^2$
$=(5a-3b)^2$
So, Possible length and breadth are $(5a-3b)$ and $(5a-3b)$.

(ii) $36 s^2-49 t^2$
$=(6s)^2-(7t)^2$
$=(6s+7t)(6s-7t)$
So, Possible length and breadth are $(6s+7t)$ and $(6s-7t)$

Question 6. Find possible expressions for the length, breadth, and heights of each of the following cuboids whose volumes are given by the following expressions in cubic units.
(i) $6 a^2-24 b^2$
(ii) $3 p s^2-15 p s+12 p$

$\textbf{Answer:}$

(i) $6 a^2-24 b^2$
$=6\left(a^2-4 b^2\right) $
$ =6(a-2 b)(a+2 b)$
So, possible dimensions of the cuboid are $6, (a-2 b)$ and $(a+2 b)$.

(ii) $3 p s^2-15 p s+12 p$
$\begin{aligned} & =3 p\left(s^2-5 s+4\right) \\ & =3 p(s-1)(s-4)\end{aligned}$
​​​​​​​So, possible dimensions of the cuboid are $3, p(s-1)$ and $(s-4)$.

Question 7. The village playground is shaped as a square of side 40 metres. A path of width $s$ metres is created around the playground for people to walk. Find an expression for the area of the path in terms of $s$.

$\textbf{Answer:}$
The village playground is shaped as a square of side 40 metres.
So, the area of the playground is $40^2=1600$ square metres
A path of width $s$ metres is created around the playground for people to walk.
So, outer side length $=(40+s)$
Outer area of the playground with the path $=(40+s)^2$
So, the area of the path
= Outer area of the playground with the path - area of the playground
$ =(40+s)^2-1600$
$=1600+160 s+4 s^2-1600 $
$=4 s^2+160 s$

Hence, the area of the path is $(4 s^2+160 s)$.

Question 8. If a number plus its reciprocal equals $\frac{10}{3}$, find the number.

$\textbf{Answer:}$
Let the number be $x$.
According to the question,
$x+\frac1x=\frac{10}3$
$⇒\frac{x^2+1}x=\frac{10}3$
$⇒3 x^2+3=10 x $
$⇒3 x^2-10 x+3=0$
$⇒3 x^2-9 x-x+3=0 $
$⇒ 3 x(x-3)-1(x-3)=0 $
$ ⇒(3 x-1)(x-3)=0$
$⇒x=\frac13,3$

Hence, the number is either $\frac13$ or 3.

Question 9. A rectangular pool has area $2 x^2+7 x+3$ square hastas. If its width is $2 x+1$ hastas, find its length. Hasta was a unit used to measure length.

$\textbf{Answer:}$
We know, Area of a rectangle = Length × width
$⇒2 x^2+7 x+3=$ Length × $2 x+1$
$\therefore$ Length $=\frac{2 x^2+7 x+3}{2 x+1}=\frac{(2 x+1)(x+3)}{2 x+1}=(x+3)$

So, the length is $(x+3)$ hastas.

*Question 10. If both $x-2$ and $x-\frac{1}{2}$ are factors of $p x^2+5 x+r$, show that $p=r$.

$\textbf{Answer:}$
$p x^2+5 x+r=(x-2)\left(x-\frac{1}{2}\right) k$
Since the polynomial is quadratic, let the leading coefficient be $p$.
$p x^2+5 x+r=p(x-2)\left(x-\frac{1}{2}\right) $
$=p\left(x^2-\frac{x}{2}-2 x+1\right) $
$ =p\left(x^2-\frac{5 x}{2}+1\right) $
$=p x^2-\frac{5 p}{2} x+p$
Comparing with $p x^2+5 x+r$, we get $r=p$

Hence, $p=r$.

*Question 11. If $a+b+c=5$ and $a b+b c+c a=10$, then prove that $a^3+b^3+c^3-3 a b c=-25$.

$\textbf{Answer:}$
Given: $a+b+c=5$ and $a b+b c+c a=10$
We know that ​​​​​​​$a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
Also,
$a^2+b^2+c^2$
$=(a+b+c)^2-2(a b+b c+c a) =5^2-2(10) =25-20 =5$
So, $a^2+b^2+c^2-a b-b c-c a=5-10=-5$
Therefore, $a^3+b^3+c^3-3 a b c=(5)(-5)=-25$

Hence proved.

*Question 12. By factoring the expression, check that $n^3-n$ is always divisible by 6 for all natural numbers $n$. Give reasons.

$\textbf{Answer:}$

Given: $n^3-n$
$
\begin{aligned}
& =\mathrm{n}\left(\mathrm{n}^2-1\right) \\
& =\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)
\end{aligned}
$
So, $\mathrm{n}^3-\mathrm{n}=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}+1)$
Thus, $\mathrm{n}^3-\mathrm{n}$ is the product of three consecutive natural numbers: $(\mathrm{n}-1), \mathrm{n},(\mathrm{n}+1)$
Among any three consecutive natural numbers:

• one is always divisible by 3
• at least one is always even, so divisible by 2

Therefore, their product is always divisible by: $2 \times 3=6$
Hence, $n^3-n$ is always divisible by 6 for all natural numbers $n$.

*Question 13. Find the value of
(i) $x^3+y^3-12 x y+64$, when $x+y=-4$
(ii) $x^3-8 y^3-36 x y-216$, when $x=2 y+6$

$\textbf{Answer:}$

(i) Given: $x+y=-4$
We know that $x^3+y^3=(x+y)^3-3 x y(x+y)$
So, $ x^3+y^3 =(-4)^3-3 x y(-4) =-64+12 x y$
Now,
$x^3+y^3-12 x y+64 $
$=(-64+12 x y)-12 x y+64 $
$ =0$

(ii)

Given that $x=2 y+6$

$\Rightarrow x-2 y-6=0$

Using the algebraic identity where if $a+b+c=0$, then $a^3+b^3+c^3=3 a b c$.

Let $a=x, b=-2 y$, and $c=-6$.

$\Rightarrow x^3+(-2 y)^3+(-6)^3=3(x)(-2 y)(-6)$

$ \Rightarrow x^3-8 y^3-216=36 x y $

$ \Rightarrow x^3-8 y^3-36 x y-216=0$

Hence, the correct answer is 0.

Exploring Algebraic Identities Class 9 Chapter 4: Topics

Topics you will learn in NCERT Class 9 Maths Chapter 4 Exploring Algebraic Identities include:

• 4.1 Introduction
• 4.2 Visualising Identities
• 4.3 Factorisation of Algebraic Expressions Using Identities
• 4.4. More Identities
• 4.5 Factorisation Using Algebra Tiles
• 4.6 Factorisation Without Using Algebra Tiles
• 4.7 Finding New Identities
• 4.8 Simplifying Rational Expressions

Class 9 Maths NCERT Chapter 4 Solutions: Extra Questions

Question 1:

The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is:

$\textbf{Answer:}$

Let the consecutive numbers be ( $\mathrm{x}-1$ ), x and ( $\mathrm{x}+1$ ).
According to the question,

$
\begin{aligned}
& (x-1)^2+x^2+(x+1)^2=365 \\
& \Rightarrow\left(x^2-2 x+1\right)+x^2+\left(x^2+2 x+1\right)=365 \\
& \Rightarrow 3 x^2=365-2 \\
& \Rightarrow 3 x^2=363 \\
& \Rightarrow x^2=121 \\
& \Rightarrow x=11
\end{aligned}
$
So, the numbers are 10, 11 and 12.
The sum of the numbers $=10+11+12=33$
Hence, the correct answer is 33.

Question 2:

If ' $a$ ' and ' $b$ ' are positive integers such that $a^2-b^2=19$, then the value of ' $a$ ' is:

$\textbf{Answer:}$

$
\begin{aligned}
& a^2-b^2=19 \\
& \Rightarrow(a+b)(a-b)=19 \times 1
\end{aligned}
$
Since ' $a$ ' and ' $b$ ' are positive integers,
$
\begin{aligned}
& \Rightarrow a+b=19 \\
& \text { and } a-b=1
\end{aligned}
$
Adding equations (i) and (ii), we get, $2 \mathrm{a}=20 \Rightarrow \mathrm{a}=10$
Hence, the correct answer is 10.

Question 3:

If $x^2+\frac{1}{x^2}=7$, then what is the value of $x^3+\frac{1}{x^3}$?

$\textbf{Answer:}$

$x^2+\frac{1}{x^2}=7$
Adding 2 on both sides, we get,
⇒ $(x+\frac{1}{x})^2 - 2 × x × \frac{1}{x} = 7+2$
⇒ $(x+\frac{1}{x})^2 = 9$
⇒ $x+\frac{1}{x} = 3$
cubing on both sides, we get,
⇒ $(x+\frac{1}{x})^3 = 3^3$
⇒ $x^3 + \frac{1}{x^3} + 3 × x × \frac{1}{x}(x+\frac{1}{x}) = 27$
⇒ $x^3 + \frac{1}{x^3} + 3 × 3 = 27$
$\therefore x^3 + \frac{1}{x^3} =18$
Hence, the correct answer is 18.

Question 4:

If $a + b + c = 6$ and $a^2+b^2+c^2=40$, then what is the value of $a^3+b^3+c^3-3abc$?

$\textbf{Answer:}$

Given that $a + b + c = 6$ and $a^2+b^2+c^2=40$,
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ac)$.
$⇒6^2 = 40 + 2(ab+bc+ac)$
$⇒ab+bc+ac = -2$
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
$⇒a^3+b^3+c^3-3abc=6 \times (40 +2)$
$⇒a^3+b^3+c^3-3abc=6 \times 42 $
$⇒a^3+b^3+c^3-3abc=252 $
Hence, the correct answer is 252.

Question 5:

What is the value of $\frac{(a+b)^2-(a-b)^2}{a b}$?

$\textbf{Answer:}$

$(a + b)^2 - (a - b)^2$
$= (a^2 + 2ab + b^2) - (a^2 – 2ab + b^2)$
$= 4ab$
So, $\frac{(a+b)^2-(a-b)^2}{a b}$
$=\frac{4ab}{ab}$
$=4$
Hence, the correct answer is 4.

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Class 9 Books & Syllabus

Before the start of a new academic year, students should refer to the latest syllabus to determine the chapters they’ll be studying. Below are the updated syllabus links, along with some recommended reference books.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9