# RD Sharma Class 9 Algebraic Identities Exercise 4.2 Solutions

On this page you will find Maths RD Sharma Class 9 Algebraic Identities Exercise 4.2 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.

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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 4.2 below and it will be beneficial for them.

## RD Sharma Class 9 Algebraic Identities Exercise 4.2 Solutions

Question 1: Write the following in the expand form:
(i): $(a+2 b+c)^{2}$
(ii): $(2 a-3 b-c)^{2}$
(iii): $(-3 x+y+z)^{2}$
(iv): $(\mathrm{m}+2 \mathrm{n}-5 \mathrm{p})^{2}$
(v): $(2+x-2 y)^{2}$
(vi): $\left(a^{2}+b^{2}+c^{2}\right)^{2}$
(vii): $(a b+b c+c a)^{2}$
(viii): $\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^{2}$
(ix): $\left(\frac{a}{b c}+\frac{b}{a c}+\frac{c}{a b}\right)^{2}$
(x): $(x+2 y+4 z)^{2}$
(xi): $(2 x-y+z)^{2}$
(xii): $(-2 x+3 y+2 z)^{2}$
Solution $1(\mathrm{i}):$
We have,
$(a+2 b+c)^{2}=a^{2}+(2 b)^{2}+c^{2}+2 a(2 b)+2 a c+2(2 b) c$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$\therefore(a+2 b+c)^{2}=a^{2}+4 b^{2}+c^{2}+4 a b+2 a c+4 b c$
Solution $1(\mathrm{ii})$ :
We have,
$(2 a-3 b-c)^{2}=[(2 a)+(-3 b)+(-c)]^{2}$
$(2 \mathrm{a})^{2}+(-3 \mathrm{~b})^{2}+(-\mathrm{c})^{2}+2(2 \mathrm{a})(-3 \mathrm{~b})+2(-3 \mathrm{~b})(-\mathrm{c})+2(2 \mathrm{a})(-\mathrm{c})$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$4 a^{2}+9 b^{2}+c^{2}-12 a b+6 b c-4 c a$
$\therefore(2 a-3 b-c)^2=4 x^2+9 y^2+c^2-12 a b+6 b c-4 c a$
Solution 1 (iii):
We have,
$(-3 x+y+z)^{2}=\left[(-3 x)^{2}+y^{2}+z^{2}+2(-3 x) y+2 y z+2(-3 x) z\right.$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$9 x^{2}+y^{2}+z^{2}-6 x y+2 y z-6 x z$
$(-3 x+y+z)^{2}=9 x^{2}+y^{2}+z^{2}-6 x y+2 x y-6 x y$
Solution 1 (iv):
We have,
$(m+2 n-5 p)^{2}=m^{2}+(2 n)^{2}+(-5 p)^{2}+2 m \times 2 n+(2 \times 2 n \times-5 p)+2 m \times-5 p$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$(m+2 n-5 p)^{2}=m^{2}+4 n^{2}+25 p^{2}+4 m n-20 n p-10 p m$
Solution $1(\mathrm{v})$
We have,
$(2+x-2 y)^{2}=2^{2}+x^{2}+(-2 y)^{2}+2(2)(x)+2(x)(-2 y)+2(2)(-2 y)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$=4+x^{2}+4 y^{2}+4 x-4 x y-8 y$
$(2+x-2 y)^{2}=4+x^{2}+4 y^{2}+4 x-4 x y-8 y$
Solution $1(\mathrm{vi}):$
We have,
$\left(a^{2}+b^{2}+c^{2}\right)^{2}=\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}+\left(c^{2}\right)^{2}+2 a^{2} b^{2}+2 b^{2} c^{2}+2 a^{2} c^{2}$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$\left(\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)^{2}=\mathrm{a}^{4}+\mathrm{b}^{4}+\mathrm{c}^{4}+2 \mathrm{a}^{2} \mathrm{~b}^{2}+2 \mathrm{~b}^{2} \mathrm{c}^{2}+2 \mathrm{c}^{2} \mathrm{a}^{2}$
Solution $1(\mathrm{vii})$ :
We have,
$(a b+b c+c a)^{2}=(a b)^{2}+(b c)^{2}+(c a)^{2}+2(a b)(b c)+2(b c)(c a)+2(a b)(c a)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$=a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2(a c) b^{2}+2(a b)(c)^{2}+2(b c)(a)^{2}$
$(a b+b c+c a)^{2}=a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+2 a c b^{2}+2 a b c^{2}+2 b c a^{2}$
Solution $1($ viii):
We have,
$\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^{2}=\left(\frac{x}{y}\right)^{2}+\left(\frac{y}{z}\right)^{2}+\left(\frac{z}{x}\right)^{2}+2 \frac{x}{y} \frac{y}{z}+2 \frac{y}{z} \frac{z}{x}+2 \frac{x}{x} \frac{x}{y}$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$\therefore\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)^{2}=\left(\frac{x^{2}}{y^{2}}\right)+\left(\frac{y^{2}}{z^{2}}\right)+\left(\frac{z^{2}}{x^{2}}\right)+2 \frac{x}{z}+2 \frac{y}{x}+2 \frac{x}{y}$
Solution $1(\mathrm{ix}):$
We have,
$\left(\frac{a}{b c}+\frac{b}{c a}+\frac{c}{a b}\right)^{2}=\left(\frac{a}{b c}\right)^{2}+\left(\frac{b}{c a}\right)^{2}+\left(\frac{c}{a b}\right)^{2}$ $+2\left(\frac{a}{b c}\right)\left(\frac{b}{c a}\right)+2\left(\frac{b}{c a}\right)\left(\frac{c}{a b}\right)+2\left(\frac{a}{b c}\right)\left(\frac{c}{a b}\right)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$\left(\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}+\frac{\mathrm{c}}{\mathrm{ab}}\right)^{2}=\left(\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2} \mathrm{c}^{2}}\right)+\left(\frac{\mathrm{b}^{2}}{\mathrm{c}^{2} \mathrm{a}^{2}}\right)+\left(\frac{\mathrm{c}^{2}}{\mathrm{a}^{2} \mathrm{~b}^{2}}\right)+\frac{2}{\mathrm{a}^{2}}+\frac{2}{\mathrm{~b}^{2}}+\frac{2}{\mathrm{c}^{2}}$
Solution $1(\mathrm{x})$ :
We have,
$(x+2 y+4 z)^{2}=x^{2}+(2 y)^{2}+(4 z)^{2}$ $+2 x \times 2 y+2 \times 2 y \times 4 z+2 x \times 4 z$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$(x+2 y+4 z)^{2}=x^{2}+4 y^{2}+16 z^{2}+4 x y+16 y z+8 x z$
Solution $1(\mathrm{xi}):$
We have,
$(2 x-y+z)^{2}=(2 x)^{2}+(-y)^{2}+(z)^{2}$ $+2(2 x)(-y)+2(-y)(z)+2(2 x)(z)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$(2 x-y+z)^{2}=4 x^{2}+y^{2}+z^{2}-4 x y-2 y z+4 x z$
Solution 1 (xii):
We have,
$(-2 x+3 y+2 z)^{2}=(-2 x)^{2}+(3 y)^{2}+(2 z)^{2}$ $+2(-2 x)(3 y)+2(3 y)(2 z)+2(-2 x)(2 z)$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$(-4 x+6 y+4 z)^{2}=4 x^{2}+9 y^{2}+4 z^{2}-12 x y+12 y z-8 x z$

Question 2: Use algebraic identities to expand the following algebraic equations.
Q 2.1: $(a+b+c)^{2}+(a-b+c)^{2}$
Ans: We have,
$(a+b+c)^{2}+(a-b+c)^{2}$ $=\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)$ $+\left(a^{2}+(-b)^{2}+c^{2}-2 a b-2 b c+2 c a\right)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$=2 a^{2}+2 b^{2}+2 c^{2}+4 c a$
$(a+b+c)^{2}+(a-b+c)^{2}=2 a^{2}+2 b^{2}+2 c^{2}+4 c a$

Q 2.2: $(a+b+c)^{2}-(a-b+c)^{2}$
Ans: We have,
$(a+b+c)^{2}-(a-b+c)^{2}=\left(a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right)$ $-\left(a^{2}+(-b)^{2}+c^{2}-2 a b-2 b c+2 c a\right)$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
$\left.=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}+2 a b+2 b c-2 c a\right)$
$=4 a b+4 b c$
$(a+b+c)^{2}-(a-b+c)^{2}=4 a b+4 b c$

Q 2.3: $(a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}$
Ans: We have,
$(a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}$ $=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$ $+\left(a^{2}+b^{2}+(z)^{2}-2 b c-2 a b+2 c a\right)$ $+\left(a^{2}+b^{2}+c^{2}-2 c a-2 b c+2 a b\right)$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$=3 \mathrm{a}^{2}+3 \mathrm{~b}^{2}+3 \mathrm{c}^{2}+2 \mathrm{ab}+2 \mathrm{bc}+2 \mathrm{ca}$ $-2 b c-2 a b-2 c a-2 b c+2 a b$
$=3 x^{2}+3 y^{2}+3 z^{2}+2 a b-2 b c+2 c a$
$(a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}$ $=3 a^{2}+3 b^{2}+3 c^{2}+2 a b-2 b c+2 c a$
$(a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}$ $=3\left(a^{2}+b^{2}+c^{2)}+2(a b-b c+c a)\right.$

Q2.4: $(2 x+p-c)^{2}-(2 x-p+c)^{2}$
Ans: We have,
$(2 x+p-c)^{2}-(2 x-p+c)^{2}$ $=\left[2 x^{2}+p^{2}+(-c)^{2}+2(2 x) p+2 p(-c)+2(2 x)(-c)\right]$ $-\left[4 x^{2}+(-p)^{2}+c^{2}+2(2 x)(-p)+2 c(-p)+2(2 x) c\right]$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$(2 x+p-c)^{2}-(2 x-p+c)^{2}=\left[4 x^{2}+p^{2}+c^{2}+4 x p-2 p c-4 x c\right]$ $-\left[4 x^{2}+p^{2}+c^{2}-4 x p-2 p c+4 x c\right]$
Opening the bracket,
$(2 x+p-c)^{2}-(2 x-p+c)^{2}=4 x^{2}+p^{2}+c^{2}$ $+4 x p-2 p c-4 c x-4 x^{2}-p^{2}-c^{2}+4 x p$ $+2 p c-4 c x]$
$(2 x+p-c)^{2}-(2 x-p+c)^{2}=8 x p-8 x c$
$=8 x(p-c)$
Hence, $(2 x+p-c)^{2}-(2 x-p+c)^{2}=8 x(p-c)$

Q2.5: $\left(\mathrm{x}^{2}+\mathrm{y}^{2}+(-\mathrm{z})^{2}\right)-\left(\mathrm{x}^{2}-\mathrm{y}^{2}+\mathrm{z}^{2}\right)^{2}$
Ans: We have,
$\left(x^{2}+y^{2}+(-z)^{2}\right)^{2}-\left(x^{2}(-y)^{2}+z^{2}\right)^{2}$ =$\left[x^{4}+y^{4}+(-z)^{4}+2 x^{2} y^{2}+2 y^{2}(-z)^{2}+2 x^{2}(-z)^{2}\right]$ $-\left[x^{4}+(-y)^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}+2 x^{2} z^{2}\right]$
$\left[\because(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 x z\right]$
= $\left[x^{4}+y^{4}+(-z)^{4}+2 x^{2} y^{2}+2 y^{2}(-z)^{2}+2 x^{2}(-z)^{2}\right]$
$-\left[x^{4}+(-y)^{4}+z^{4}-2 x^{2} y^{2}-2 y^{2} z^{2}+2 x^{2} z^{2}\right]$
$=4 x^{2} y^{2}-4 z^{2} x^{2}$
Hence, $\left(x^{2}+y^{2}+(-z)^{2}\right)^{2}-\left(x^{2}(-y)^{2}+z^{2}\right)^{2}=4 x^{2} y^{2}-4 z^{2} x^{2}$

Q3: If $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=16$, find the value of $a b+b c+c a:$
Ans: We know that,
$\left[\because(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a\right]$
$(0)^{2}=16+2(a b+b c+c a)$
$2(a b+b c+c a)=-16$
$a b+b c+c a=-8$
Hence, value of required express $a b+b c+c a=-8$

Q4: If $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=16$ and $a b+b c+c a=10$, find the value of $a+b+c$ ?
Ans: We know that,
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$(x+y+z)^{2}=16+2(10)$
$(x+y+z)^{2}=36$
$(x+y+z)=\sqrt{36}$
$(x+y+z)=\pm 6$
Hence, value of required expression $\mathrm{I} ;(\mathrm{a}+\mathrm{b}+\mathrm{c})=\pm 8$

Q5: If $a+b+c=9$ and $a b+b c+c a=23$, find value of $a^{2}+b^{2}+c^{2}$
Ans: We know that,
$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)$
$9^{2}=a^{2}+b^{2}+c^{2}+2(23)$
$81=a^{2}+b^{2}+c^{2}+46$
$a^{2}+b^{2}+c^{2}=81-46$
$a^{2}+b^{2}+c^{2}=35$
Hence, value of required expression $\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=35$

Q6: Find the value of the equation : $4 \mathrm{x}^{2}+\mathrm{y}^{2}+25 \mathrm{z}^{2}+4 \mathrm{xy}-10 \mathrm{yz}-20 \mathrm{zx}$ when $x=4, \mathrm{y}=3, \mathrm{z}=2$
Ans: $4 x^{2}+y^{2}+25 z^{2}+4 x y-10 y z-20 z x$
$(2 x)^{2}+y^{2}+(-5 z)^{2}+2(2 x)(y)+2(y)(-5 z)+2(-5 z)(2 x)$
$(2 x+y-5 z)^{2}$
$(2(4)+3-5(2))^{2}$
$(8+3-10)^{2}$
$(1)^{2}$
1
Hence value of the equation is equals to 1

Q7: Simplify each of the following expressions:
Q 7.1: $(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}+\left(\mathrm{x}+\frac{\mathrm{y}}{2}+\frac{2}{3}\right)^{2}-\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{y}}{3}+\frac{\mathrm{z}}{4}\right)^{2}$
Ans: Expanding, we get
$=\left[x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x\right]$ $+\left[\mathrm{x}^{2}+\frac{\mathrm{y}^{2}}{4}+\frac{\mathrm{z}^{2}}{9}+2 \mathrm{x} \frac{\mathrm{y}}{2}+2 \frac{\mathrm{zx}}{3}+\frac{\mathrm{yz}}{3}\right]$ $-\left[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{z^{2}}{10}+\frac{x y}{3}+\frac{y z}{6}+\frac{x z}{4}\right]$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$=x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x+x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}$ $+2 x \frac{y}{2}+\frac{x y}{3}+\frac{2 z x}{3}-\frac{x^{2}}{4}-\frac{y^{2}}{9}-\frac{z^{2}}{10}-\frac{x y}{3}$ $-\frac{\mathrm{yZ}}{6}-\frac{\mathrm{XZ}}{4}$
Rearranging coefficients,
$=\frac{8 x^{2}-x^{2}}{4}+\frac{36 y^{2}+9 y^{2}-4 y^{2}}{36}$ $+\frac{144 z^{2}+16 z^{2}-9 z^{2}}{144}+\frac{6 x y+3 x y-x y}{3}+\frac{13 y z}{6}+\frac{29 x z}{12}$
$=\frac{7 x^{2}}{4}+\frac{41 y^{2}}{36}+\frac{151 z^{2}}{144}+\frac{8 x y}{3}+\frac{13 y z}{6}+\frac{29 z x}{12}$
$(x+y+z)^{2}+\left(x+\frac{y}{2}+\frac{z}{3}\right)^{2}-\left(\frac{x}{2}+\frac{y}{3}+\frac{z}{4}\right)^{2}$ $=\frac{7 x^{2}}{4}+\frac{41 y^{2}}{36}+\frac{151 z^{2}}{144}+\frac{8 x y}{3}+\frac{13 y z}{6}+\frac{29 z x}{12}$

Q7.2: $(\mathrm{x}+\mathrm{y}-2 \mathrm{z})^{2}-\mathrm{x}^{2}-\mathrm{y}^{2}-3 \mathrm{z}^{2}+4 \mathrm{xy}$
Ans: $\quad(\mathrm{x}+\mathrm{y}-2 \mathrm{z})^{2}-\mathrm{x}^{2}-\mathrm{y}^{2}-3 \mathrm{z}^{2}+4 \mathrm{xy}$
$=\left[x^{2}+y^{2}+4 z^{2}+2 x y+2 y(-2 z)\right.$ $+2 a(-2 c)]-x^{2}-y^{2}-3 z^{2}+4 x y$
$=z^{2}+6 x y-4 y z-4 z x$
$(x+y-2 z)^{2}-x^{2}-y^{2}-3 z^{2}+4 x y=z^{2}+6 x y-4 y z-4 z x$

Q7.3: $\left[x^{2}-x+1\right]^{2}-\left[x^{2}+x+1\right]^{2}$
Ans: $\left[\mathrm{x}^{2}-\mathrm{x}+1\right]^{2}-\left[\mathrm{x}^{2}+\mathrm{x}+1\right]^{2}$
$\left.=\left(\mathrm{x}^{2}\right)^{2}+(-\mathrm{x})^{2}+1^{2}+2\left(\mathrm{x}^{2}\right)(-\mathrm{x})+2(-\mathrm{x})(1)+2 \mathrm{x}^{2}\right)$ $-\left[\left(x^{2}\right)^{2}+x^{2}+1+2 x^{2} x+2 x(1)+2 x^{2}(1)\right]$
$\left[\because(\mathrm{x}+\mathrm{y}+\mathrm{z})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2 \mathrm{xy}+2 \mathrm{yz}+2 \mathrm{xz}\right]$
$=x^{4}+y^{2}+1-2 x^{3}-2 x+2 x^{2}-x^{2}-x^{4}-1-2 x^{3}-2 x-2 x^{2}$
$=-4 x^{3}-4 x$
$=-4 x\left(x^{2}+1\right)$
Hence simplified equation $=\left[\mathrm{x}^{2}-\mathrm{x}+1\right]^{2}-\left[\mathrm{x}^{2}+\mathrm{x}+1\right]^{2}=-4 \mathrm{x}\left(\mathrm{x}^{2}+1\right)$

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