# RD Sharma Class 9 Algebraic Identities Exercise 4.1 Solutions

On this page you will find Maths RD Sharma Class 9 Algebraic Identities Exercise 4.1 Solutions. The solutions provided here are to help students practice math problems and get better at solving difficult chapter questions. Maths chapter 4 for class 9 deals with the topic of triangles and it is one of the most important chapters.

## Download RD Sharma Class 9 Algebraic Identities Exercise 4.1 Solutions in PDF

Students can use these solutions to overcome the fear of maths and the solutions have been designed in such a way that it enables them to discover easy ways to solve different problems. These solutions can help students in refining their maths fluency and problem-solving skills. Students can go through the RD Sharma solutions for class 9 chapter 4 Algebraic Identities Exercise 4.1 below and it will be beneficial for them.

## RD Sharma Class 9 Algebraic Identities Exercise 4.1 Solutions

1) Evaluate each of the following using identities:
(i) $\left(2 x-\frac{1}{x}\right)^{2}$
Solution:
Given,
$\left(2 x-\frac{1}{x}\right)^{2}=(2 x)^{2}+\left(\frac{1}{x}\right)^{2}-2 * 2 x * \frac{1}{x}$
$\left(2 x-\frac{1}{x}\right)^{2}=4 x^{2}+\frac{1}{x^{2}}-4 \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
Where, $\mathrm{a}=2 \mathrm{x}, \mathrm{b}=\frac{1}{\mathrm{x}}$
$\therefore\left(2 \mathrm{x}-\frac{1}{\mathrm{x}}\right)^{2}=4 \mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}-4$
(ii) $(2 x+y)(2 x-y)$
Solution:
Given, $(2 x+y)(2 x-y)$
$=(2 x)^{2}-(y)^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$=4 x^{2}-y^{2}$
$\therefore(2 x+y)(2 x-y)=4 x^{2}-y^{2}$
(iii) $\left(a^{2} b-a b^{2}\right)^{2}$
Solution:
Given, $\left(a^{2} b-a b^{2}\right)^{2}$
$=\left(\mathrm{a}^{2} \mathrm{~b}\right)^{2}+\left(\mathrm{ab}^{2}\right)^{2}-2 * \mathrm{a}^{2} \mathrm{~b} * \mathrm{ab}^{2} \quad\left[\because(\mathrm{a}-\mathrm{b})^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab}\right]$
Where, $a=a^{2} b, b=a b^{2}$
$=a^{4} b^{2}+b^{4} a^{2}-2 a^{3} b^{3}$
$\therefore\left(a^{2} b-a b^{2}\right)^{2}=a^{4} b^{2}+b^{4} a^{2}-2 a^{3} b^{3}$
(iv) $(a-0.1)(a+0.1)$
Solution:
Given, $(a-0.1)(a+0.1)$
$=a^{2}-(0.1)^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
Where, $a=a$ and $b=0.1$
$=a^{2}-0.01$
$\therefore(\mathrm{a}-0.1)(\mathrm{a}+0.1)=\mathrm{a}^{2}-0.01$
(v) $\left(1.5 x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)$
Solution:
Given, $\left(1.5 x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)$
$=\left(1.5 x^{2}\right)^{2}-\left(0.3 y^{2}\right)^{2} \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
Where, $a=1.5 x^{2}, b=0.3 y^{2}$
$=2.25 x^{4}-0.09 y^{4}$
$\therefore\left(1.5 x^{2}-0.3 y^{2}\right)\left(1.5 x^{2}+0.3 y^{2}\right)=2.25 x^{4}-0.09 y^{4}$

2) Evaluate each of the following using identities:
(i) $(399)^{2}$
Solution:
We have,
$399^{2}=(400-1)^{2}$
$=(400)^{2}+(1)^{2}-2 \times 400 \times 1 \quad\left[(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
Where, $a=400$ and $b=1$
$=160000+1-8000$
$=159201$
Therefore, $(399)^{2}=159201$.
(ii) $(0.98)^{2}$
Solution:
We have,
$(0.98)^{2}=(1-0.02)^{2}$
$=(1)^{2}+(0.02)^{2}-2 \times 1 \times 0.02$
$=1+0.0004-0.04$
[Where, $a=1$ and $b=0.02$ ]
$=1.0004-0.04$
$=0.9604$
Therefore, $(0.98)^{2}=0.9604$
(iii) $991 \times 1009$
Solution:
We have,
$991 \times 1009$
$=(1000-9)(1000+9)$
$\begin{array}{ll}=(1000)^{2}-(9)^{2} & {\left[(a+b)(a-b)=a^{2}-b^{2}\right]} \\ =1000000-81 & {[\text { Where } a=1000 \text { and } b=9]}\end{array}$
$=999919$
Therefore, $991 \times 1009=999919$
(iv) $117 \times 83$
Solution:
We have,
$117 \times 83$
$=(100+17)(100-17)$
$=(100)^{2}-(17)^{2} \quad\left[(a+b)(a-b)=a^{2}-b^{2}\right]$
$=10000-289 \quad[$ Where $a=100$ and $b=17]$
$=9711$
Therefore, $117 \times 83=9711$

3) Simplify each of the following:
(i) $175 \times 175+2 \times 175 \times 25+25 \times 25$
Solution:
We have,
$175 \times 175+2 \times 175 \times 25+25 \times 25=(175)^{2}+2(175)(25)+(25)^{2}$
$\begin{array}{ll}=(175+25)^{2} & {\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right]} \\ =(200)^{2} & {[\text { Where } a=175 \text { and } b=25]}\end{array}$
$=40000$
Therefore, $175 \times 175+2 \times 175 \times 25+25 \times 25=40000$
(ii) $322 \times 322-2 \times 322 \times 22+22 \times 22$
Solution:
We have,
$322 \times 322-2 \times 322 \times 22+22 \times 22$
$\begin{array}{ll}=(322-22)^{2} & {\left[a^{2}+b^{2}-2 a b=(a-b)^{2}\right]} \\ =(300)^{2} & {[\text { Where } a=322 \text { and } b=22]}\end{array}$
$=90000$
Therefore, $322 \times 322-2 \times 322 \times 22+22 \times 22=90000$.
(iii) $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$
Solution:
We have,
$0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24$
$\begin{array}{ll}=(0.76+0.24)^{2} & {\left[a^{2}+b^{2}+2 a b=(a+b)^{2}\right]} \\ =(1.00)^{2} & {[\text { Where } a=0.76 \text { and } b=0.24]}\end{array}$
$=1$
Therefore, $0.76 \times 0.76+2 \times 0.76 \times 0.24+0.24 \times 0.24=1$
(iv) $\frac{7.83 * 7.83-1.17 * 1.17}{6.66}$
Solution:
We have,
$\frac{7.83 * 7.83-1.17 * 1.17}{6.66}$
$=\frac{(7.83+1.17)(7.83-1.17)}{6.66} \quad\left[\because(a-b)^{2}=(a+b)(a-b)\right]$
$=\frac{(9.00)(6.66)}{(6.66)}$
$=9$
$\therefore \frac{7.83 * 7.83-1.17 * 1.17}{6.66}=9$

4) If $x+\frac{1}{x}=11$, find the value of $x^{2}+\frac{1}{x^{2}}$
Solution:
We have, $\mathrm{x}+\frac{1}{\mathrm{x}}=11$
Now, $\left(x+\frac{1}{x}\right)^{2}=x^{2}+\left(\frac{1}{x}\right)^{2}+2 * x * \frac{1}{x}$
$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$
$\Rightarrow(11)^{2}=x^{2}+\frac{1}{x^{2}}+2 \quad\left[\because x+\frac{1}{x}=11\right]$
$\Rightarrow 121=x^{2}+\frac{1}{x^{2}}+2$
$\Rightarrow \mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}=119$

5) If $x-\frac{1}{x}=-1$, find the value of $x^{2}+\frac{1}{x^{2}}$
Solution:
We have, $\mathrm{x}-\frac{1}{\mathrm{x}}=-1$
Now, $\left(x-\frac{1}{x}\right)^{2}=x^{2}+\left(\frac{1}{x}\right)^{2}-2 * x * \frac{1}{x}$
$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-2$
$\Rightarrow(-1)^{2}=\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}-2 \quad\left[\because \mathrm{x}-\frac{1}{\mathrm{x}}=-1\right]$
$\Rightarrow 2+1=x^{2}+\frac{1}{x^{2}}$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=3$

6) If $x+\frac{1}{x}=\sqrt{5}$, find the value of $x^{2}+\frac{1}{x^{2}}$ and $x^{4}+\frac{1}{x^{4}}$
Solution:
We have,
$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\left(\frac{1}{x}\right)^{2}+2 * x * \frac{1}{x}$
$\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2}=\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}+2$
$\Rightarrow(\sqrt{5})^{2}=x^{2}+\frac{1}{x^{2}}+2 \quad\left[\because x+\frac{1}{x}=\sqrt{5}\right]$
$\Rightarrow 5=x^{2}+\frac{1}{x^{2}}+2$
$\Rightarrow x^{2}+\frac{1}{x^{2}}=3$ ……(1)
Now, $\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=x^{4}+\frac{1}{x^{4}}+2 * x^{2} * \frac{1}{x^{2}}$
$\Rightarrow\left(\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}\right)^{2}=\mathrm{x}^{4}+\frac{1}{\mathrm{x}^{4}}+2$
$\Rightarrow 9=x^{4}+\frac{1}{x^{4}}+2$ $\left[\because \mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}=3\right]$
$\Rightarrow x^{4}+\frac{1}{x^{4}}=7$
Hence, $\mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}=3 ; \mathrm{x}^{4}+\frac{1}{\mathrm{x}^{4}}=7$

7) If $x^{2}+\frac{1}{x^{2}}=66$, find the value of $x-\frac{1}{x}$
Solution:
We have,
$\left(x-\frac{1}{x}\right)^{2}=x^{2}+\left(\frac{1}{x}\right)^{2}-2 * x * \frac{1}{x}$
$\Rightarrow\left(x-\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}-2$
$\Rightarrow\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{2}=66-2 \quad\left[\because \mathrm{x}^{2}+\frac{1}{\mathrm{x}^{2}}=66\right]$
$\Rightarrow\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{2}=64$
$\Rightarrow\left(\mathrm{x}-\frac{1}{\mathrm{x}}\right)^{2}=(\pm 8)^{2}$
$\Rightarrow \mathrm{x}-\frac{1}{\mathrm{x}}=\pm 8$

8) If $x^{2}+\frac{1}{x^{2}}=79$, find the value of $x+\frac{1}{x}$
Solution:
We have,
$\left(x+\frac{1}{x}\right)^{2}=x^{2}+\left(\frac{1}{x}\right)^{2}+2 * x * \frac{1}{x}$
$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=x^{2}+\frac{1}{x^{2}}+2$
$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=79+2 \quad\left[\because x^{2}+\frac{1}{x^{2}}=79\right]$
$\Rightarrow\left(x+\frac{1}{x}\right)^{2}=81$
$\Rightarrow\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)^{2}=(\pm 9)^{2}$
$\Rightarrow \mathrm{x}+\frac{1}{\mathrm{x}}=\pm 9$

9) If $9 x^{2}+25 y^{2}=181$ and $x y=-6$, find the value of $3 x+5 y$.
Solution:
We have,
$(3 x+5 y)^{2}=(3 x)^{2}+(5 y)^{2}+2 * 3 x * 5 y$
$\Rightarrow(3 x+5 y)^{2}=9 x^{2}+25 y^{2}+30 x y$
$=181+30(-6) \quad\left[\right.$ Since, $9 x^{2}+25 y^{2}=181$ and $\left.x y=-6\right]$
$\Rightarrow(3 x+5 y)^{2}=1$
$\Rightarrow(3 x+5 y)^{2}=(\pm 1)^{2}$
$\Rightarrow 3 x+5 y=\pm 1$

10) If $2 x+3 y=8$ and $x y=2$, find the value of $4 x^{2}+9 y^{2}$
Solution:
We have,
$(2 x+3 y)^{2}=(2 x)^{2}+(3 y)^{2}+2 * 2 x^{\star} 3 y$
$\Rightarrow(2 x+3 y)^{2}=4 x^{2}+9 y^{2}+12 x y$ [Since, $2 x+3 y=8$ and $x y=24]$
$\Rightarrow(8)^{2}=4 x^{2}+9 y^{2}+24$
$\Rightarrow 64-24=4 x^{2}+9 y^{2}$
$\Rightarrow 4 x^{2}+9 y^{2}=40$

11) If $3 x-7 y=10$ and $x y=-1$, find the value of $9 x^{2}+49 y^{2}$.
Solution:
We have,
$(2-7 y)^{2}=(3 x)^{2}+(-7 y)^{2}-2 * 3 x^{*} 7 y$
$\Rightarrow(3 x-7 y)^{2}=9 x^{2}+49 y^{2}-42 x y$ $[$ Since, $3 x-7 y=10$ and $x y=-1]$
$\Rightarrow(10)^{2}=9 x^{2}+49 y^{2}+42$
$\Rightarrow 100-42=9 x^{2}+49 y^{2}$
$\Rightarrow 9 x^{2}+49 y^{2}=58$

12) Simplify each of the following products:
(i) $\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)$
Solution:
$\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)$
$\left.\Rightarrow\left[\left(\frac{1}{2} a\right)^{2}-(3 b)^{2}\right)\right]\left[\left(\frac{1}{4} a^{2}+9 b^{2}\right)\right]$ $\left.\left[\because(a+b)(a-b)=a^{2}-b^{2}\right)\right]$
$\left.\Rightarrow\left[\frac{1}{4} a^{2}-9 b^{2}\right)\right]\left[\frac{1}{4} a^{2}+9 b^{2}\right]$ $\left.\left[\because(a b)^{2}=a^{2} b^{2}\right)\right]$
$=\left[\left(\frac{1}{4} a^{2}\right)^{2}-\left(9 b^{2}\right)^{2}\right]$ $\left.\left[\because(a+b)(a-b)=a^{2}-b^{2}\right)\right]$
$=\frac{1}{16} a^{4}-81 b^{4}$
$\therefore\left(\frac{1}{2} a-3 b\right)\left(3 b+\frac{1}{2} a\right)\left(\frac{1}{4} a^{2}+9 b^{2}\right)=\frac{1}{16} a^{4}-81 b^{4}$
(ii) $\left(\mathrm{m}+\frac{\mathrm{n}}{7}\right)^{3}\left(\mathrm{~m}-\frac{\mathrm{n}}{7}\right)$
Solution:
We have,
$\left(m+\frac{n}{7}\right)^{3}\left(m-\frac{n}{7}\right)$
$=\left(m+\frac{n}{7}\right)\left(m+\frac{n}{7}\right)\left(m+\frac{n}{7}\right)\left(m-\frac{n}{7}\right)$
$=\left(m+\frac{n}{7}\right)^{2}\left(m^{2}-\left(\frac{n}{7}\right)^{2}\right)$ $\left[\because(a+b)(a+b)=(a+b)^{2}\right.$ and $\left.(a+b)(a-b)=a^{2}-b^{2}\right]$
$=\left(m+\frac{n}{7}\right)^{2}\left(m^{2}-\frac{n^{2}}{49}\right)$
$\therefore\left(m+\frac{n}{7}\right)^{3}\left(m-\frac{n}{7}\right)=\left(m+\frac{n}{7}\right)^{2}\left(m^{2}-\frac{n^{2}}{49}\right)$
(iii) $\left(\frac{x}{2}-\frac{2}{5}\right)\left(\frac{2}{5}-\frac{x}{2}\right)-x^{2}+2 x$
Solution:
We have,
$\left(\frac{x}{2}-\frac{2}{5}\right)\left(\frac{2}{5}-\frac{x}{2}\right)-x^{2}+2 x$
$\Rightarrow-\left(\frac{2}{5}-\frac{x}{2}\right)\left(\frac{2}{5}-\frac{x}{2}\right)-x^{2}+2 x$
$\Rightarrow-\left(\frac{2}{5}-\frac{x}{2}\right)^{2}-x^{2}+2 x \quad\left[\because(a-b)(a-b)=(a-b)^{2}\right]$
$\Rightarrow-\left[\left(\frac{2}{5}\right)^{2}+\left(\frac{x}{2}\right)^{2}-2\left(\frac{2}{5}\right)\left(\frac{x}{2}\right)\right]-x^{2}+2 x$
$\Rightarrow-\left(\frac{4}{25}+\frac{x^{2}}{4}-\frac{2 x}{5}\right)-x^{2}+2 x$
$\Rightarrow-\frac{x^{2}}{4}-x^{2}+\frac{2 x}{5}+2 x-\frac{4}{25}$
$\Rightarrow-\frac{5 x^{2}}{4}+\frac{12 x}{5}-\frac{4}{25}$
$\therefore\left(\frac{x}{2}-\frac{2}{5}\right)\left(\frac{2}{5}-\frac{x}{2}\right)-x^{2}+2 x=-\frac{5 x^{2}}{4}+\frac{12 x}{5}-\frac{4}{25}$
(iv) $\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)$
Solution:
$\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)$
$\left[(x)^{2}+(x-2)\right]\left[\left(x^{2}-(x+2)\right]\right.$
$\Rightarrow\left(x^{2}\right)^{2}-(x-2)^{2}$ $\left[(a-b)(a+b)=a^{2}-b^{2}\right]$
$\Rightarrow x^{4}-\left(x^{2}+4-4 x\right)$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\Rightarrow x^{4}-x^{2}+4 x-4$
$\therefore\left(x^{2}+x-2\right)\left(x^{2}-x+2\right)=x^{4}-x^{2}+4 x-4$
(v) $\left(x^{3}-3 x-x\right)\left(x^{2}-3 x+1\right)$
Solution:
We have,
$\left(x^{3}-3 x-x\right)\left(x^{2}-3 x+1\right)$
$\Rightarrow x\left(x^{2}-3 x-1\right)\left(x^{2}-3 x+1\right)$
$\Rightarrow \mathrm{X}\left[\left(\mathrm{x}^{2}-3 \mathrm{x}\right)^{2}-(1)^{2}\right]$ $\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$\Rightarrow \mathrm{x}\left[\left(\mathrm{x}^{2}\right)^{2}+(-3 \mathrm{x})^{2}-2(3 \mathrm{x})\left(\mathrm{x}^{2}\right)-1\right]$
$\Rightarrow \mathrm{x}\left[\mathrm{x}^{4}+9 \mathrm{x}^{2}-6 \mathrm{x}^{3}-1\right]$
$\Rightarrow x^{5}-6 x^{4}+9 x^{3}-x$
$\therefore\left(x^{3}-3 x-x\right)\left(x^{2}-3 x+1\right)=x^{5}-6 x^{4}+9 x^{3}-x$
(vi) $\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)$
Solution:
We have,
$\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)$
$\Rightarrow\left[\left(2 x^{4}-4 x^{2}\right)^{2}-(1)^{2}\right] \quad\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$
$\Rightarrow\left[\left(2 x^{4}\right)^{2}+\left(4 x^{2}\right)^{2}-2\left(2 x^{4}\right)\left(4 x^{2}\right)-1\right]$
$\Rightarrow 4 x^{8}-16 x^{6}+16 x^{4}-1$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$\therefore\left(2 x^{4}-4 x^{2}+1\right)\left(2 x^{4}-4 x^{2}-1\right)=4 x^{8}-16 x^{6}+16 x^{4}-1$

13) Prove that $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca}$ is always non-negative for all values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$.
Solution:
We have,
$a^{2}+b^{2}+c^{2}-a b-b c-c a$
Multiply and divide by ‘2’
$=\frac{2}{2}\left[a^{2}+b^{2}+c^{2}-a b-b c-c a\right]$
$=\frac{1}{2}\left[2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right]$
$=\frac{1}{2}\left[a^{2}+a^{2}+b^{2}+b^{2}+c^{2}+c^{2}-2 a b-2 b c-2 c a\right]$
$=\frac{1}{2}\left[\left(a^{2}+b^{2}-2 a b\right)+\left(a^{2}+c^{2}-2 c a\right)+\left(b^{2}+c^{2}-2 b c\right)\right]$
$=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$ $\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$
$=\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2} \geq 0$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}-\mathrm{ab}-\mathrm{bc}-\mathrm{ca} \geq 0$
Hence, $a^{2}+b^{2}+c^{2}-a b-b c-c a \geq 0$ is always non-negative for all values of $a, b$ and $c$.

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