RD Sharma Solutions Class 9 Algebraic Identities Exercise 4.2

RD Sharma Class 9 Solutions Chapter 4 Ex 4.2 Free Download

RD Sharma Solutions Class 9 Chapter 4 Ex 4.2

Question 1: Write the following in the expand 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): \((a^{2}+b^{2}+c^{2})^{2}\)

(vii): \((ab+bc+ca)^{2}\)

(viii): \((\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{2}\)

(ix): \((\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab})^{2}\)

(x): \((x+2y+4z)^{2}\)

(xi): \((2x-y+z)^{2}\)

(xii): \((-2x+3y+2z)^{2}\)

Solution 1(i):

We have,

\((a+2b+c)^{2}=a^{2}+(2b)^{2}+c^{2}+2a(2b)+2ac+2(2b)c\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(∴ (a+2b+c)^{2}= a^{2}+4b^{2}+c^{2}+4ab+2ac+4bc\)

Solution 1(ii):

We have,

\((2a-3b-c)^{2}=[(2a)+(-3b)+(-c)]^{2}\)

\((2a)^{2}+(-3b)^{2}+(-c)^{2}+2(2a)(-3b)+2(-3b)(-c)+2(2a)(-c)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(4a^{2}+9b^{2}+c^{2}-12ab+6bc-4ca\)

∴ (2a-3b-c)^{2}=4x^{2}+9y^{2}+c^{2}-12ab+6bc-4ca

Solution 1(iii):

We have,

\((-3x+y+z)^{2}=[(-3x)^{2}+y^{2}+z^{2}+2(-3x)y+2yz+2(-3x)z\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(9x^{2}+y^{2}+z^{2}-6xy+2yz-6xz\)

\((-3x+y+z)^{2}=9x^{2}+y^{2}+z^{2}-6xy+2xy-6xy\)

Solution 1(iv):

We have,

\((m+2n-5p)^{2}=m^{2}+(2n)^{2}+(-5p)^{2}+2m\times 2n+(2\times 2n\times -5p)+2m\times -5p\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((m+2n-5p)^{2}=m^{2}+4n^{2}+25p^{2}+4mn-20np-10pm\)

Solution 1(v):

We have,

\((2+x-2y)^{2}=2^{2}+x^{2}+(-2y)^{2}+2(2)(x)+2(x)(-2y)+2(2)(-2y)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=4+x^{2}+4y^{2}+4x-4xy-8y\)

\((2+x-2y)^{2}=4+x^{2}+4y^{2}+4x-4xy-8y\)

Solution 1(vi):

We have,

\((a^{2}+b^{2}+c^{2})^{2}=(a^{2})^{2}+(b^{2})^{2}+(c^{2})^{2}+2a^{2}b^{2}+2b^{2}c^{2}+2a^{2}c^{2}\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((a^{2}+b^{2}+c^{2})^{2}=a^{4}+b^{4}+c^{4}+2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}\)

Solution 1(vii):

We have,

\((ab+bc+ca)^{2}=(ab)^{2}+(bc)^{2}+(ca)^{2}+2(ab)(bc)+2(bc)(ca)+2(ab)(ca)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2(ac)b^{2}+2(ab)(c)^{2}+2(bc)(a)^{2}\)

\((ab+bc+ca)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2acb^{2}+2abc^{2}+2bca^{2}\)

Solution 1(viii):

We have,

\((\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{2}=(\frac{x}{y})^{2}+(\frac{y}{z})^{2}+(\frac{z}{x})^{2}+2\frac{x}{y}\frac{y}{z}+2\frac{y}{z}\frac{z}{x}+2\frac{z}{x}\frac{x}{y}\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(∴ (\frac{x}{y}+\frac{y}{z}+\frac{z}{x})^{2}=(\frac{{x}^{2}}{{y}^{2}})+(\frac{{y}^{2}}{{z}^{2}})+(\frac{{z}^{2}}{{x}^{2}})+2\frac{x}{z}+2\frac{y}{x}+2\frac{x}{y}\)

Solution 1(ix):

We have,

\((\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab})^{2}=(\frac{a}{bc})^{2}+(\frac{b}{ca})^{2}+(\frac{c}{ab})^{2}+2(\frac{a}{bc})(\frac{b}{ca})+2(\frac{b}{ca})(\frac{c}{ab})+2(\frac{a}{bc})(\frac{c}{ab})\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab})^{2}=(\frac{a^{2}}{b^{2}c^{2}})+(\frac{b^{2}}{c^{2}a^{2}})+(\frac{c^{2}}{a^{2}b^{2}})+\frac{2}{a^{2}}+\frac{2}{b^{2}}+\frac{2}{c^{2}}\)

Solution 1(x):

We have,

\((x+2y+4z)^{2}=x^{2}+(2y)^{2}+(4z)^{2}+2x\times 2y+2\times 2y\times 4z+2x\times4z\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((x+2y+4z)^{2}=x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz\)

Solution 1(xi):

We have,

\((2x-y+z)^{2}=(2x)^{2}+(-y)^{2}+(z)^{2}+2(2x)(-y)+2(-y)(z)+2(2x)(z)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((2x-y+z)^{2}=4x^{2}+y^{2}+z^{2}-4xy-2yz+4xz\)

Solution 1 (xii):

We have,

\((-2x+3y+2z)^{2}=(-2x)^{2}+(3y)^{2}+(2z)^{2}+2(-2x)(3y)+2(3y)(2z)+2(-2x)(2z)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((-4x+6y+4z)^{2}=4x^{2}+9y^{2}+4z^{2}-12xy+12yz-8xz\)

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,

\((a+b+c)^{2}+(a-b+c)^{2}\) =\((a^{2}+b^{2}+c^{2}+2ab+2bc+2ca)+(a^{2}+(-b)^{2}+c^{2}-2ab-2bc+2ca)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=2a^{2}+2b^{2}+2c^{2}+4ca\)

                \((a+b+c)^{2}+(a-b+c)^{2}=2a^{2}+2b^{2}+2c^{2}+4ca\)

Q 2.2: \((a+b+c)^{2}-(a-b+c)^{2}\)

Ans:       We have,

\((a+b+c)^{2}-(a-b+c)^{2}= (a^{2}+b^{2}+c^{2}+2ab+2bc+2ca)-(a^{2}+(-b)^{2}+c^{2}-2ab-2bc+2ca)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca-a^{2}-b^{2}-c^{2}+2ab+2bc-2ca)\)

\(=4ab+4bc\)

\((a+b+c)^{2}-(a-b+c)^{2}=4ab+4bc \)

Q 2.3: \((a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}\)

Ans:       We have,

\((a+b+c)^{2}+(a+b-c)^{2}+(a+b-c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca+(a^{2}+b^{2}+(z)^{2}-2bc-2ab+2ca)+(a^{2}+b^{2}+c^{2}-2ca-2bc+2ab)\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=3a^{2}+3b^{2}+3c^{2}+2ab+2bc+2ca-2bc-2ab-2ca-2bc+2ab\)

\(=3x^{2}+3y^{2}+3z^{2}+2ab-2bc+2ca\)

\((a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}= 3a^{2}+3b^{2}+3c^{2}+2ab-2bc+2ca\)

\((a+b+c)^{2}+(a+b-c)^{2}+(a-b+c)^{2}= 3(a^{2}+b^{2}+c^{2)}+2(ab-bc+ca)\)

Q 2.4:    \((2x+p-c)^{2}-(2x-p+c)^{2}\)

Ans:      We have,

\((2x+p-c)^{2}-(2x-p+c)^{2}=[2x^{2}+p^{2}+(-c)^{2}+2(2x)p+2p(-c)+2(2x)(-c)]-[4x^{2}+(-p)^{2}+c^{2}+2(2x)(-p)+2c(-p)+2(2x)c]\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\((2x+p-c)^{2}-(2x-p+c)^{2}=[4x^{2}+p^{2}+c^{2}+4xp-2pc-4xc]-[4x^{2}+p^{2}+c^{2}-4xp-2pc+4xc]\)

Opening the bracket,

\((2x+p-c)^{2}-(2x-p+c)^{2}=4x^{2}+p^{2}+c^{2}+4xp-2pc-4cx -4x^{2}-p^{2}-c^{2}+4xp+2pc-4cx]\)

\((2x+p-c)^{2}-(2x-p+c)^{2}=8xp-8xc\)

=\(8x(p-c)\)

Hence, \((2x+p-c)^{2}-(2x-p+c)^{2}=8x(p-c)\)

Q 2.5: \((x^{2}+y^{2}+(-z)^{2})-(x^{2}-y^{2}+z^{2})^{2}\)

Ans:       We have,

\((x^{2}+y^{2}+(-z)^{2})^{2}-(x^{2} (-y)^{2}+z^{2})^{2}\)

=\([x^{4}+y^{4}+(-z)^{4}+2x^{2}y^{2}+2y^{2}(-z)^{2}+2x^{2}(-z)^{2}]-[x^{4}+(-y)^{4}+z^{4}-2x^{2}y^{2}-2y^{2}z^{2}+2x^{2}z^{2}]\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

Taking the negative sign inside,

= \([x^{4}+y^{4}+(-z)^{4}+2x^{2}y^{2}+2y^{2}(-z)^{2}+2x^{2}(-z)^{2}]-[x^{4}+(-y)^{4}+z^{4}-2x^{2}y^{2}-2y^{2}z^{2}+2x^{2}z^{2}]\)

= \(4x^{2}y^{2} – 4z^{2}x^{2}\)

Hence, \((x^{2}+y^{2}+(-z)^{2})^{2}-(x^{2} (-y)^{2}+z^{2})^{2} = 4x^{2}y^{2} – 4z^{2}x^{2}\)

Q3: If a+b+c=0 and \(a^{2}+b^{2}+c^{2}=16\), find the value of ab+bc+ca:

Ans:       We know that,

\([∵ (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca]\)

  • \((0)^{2}\)=16+2(ab+bc+ca)
  • 2(ab+bc+ca)=-16
  • ab+bc+ca=-8

Hence, value of required express ab+bc+ca =-8

Q4: If \(a^{2}+b^{2}+c^{2}=16\) and ab+bc+ca=10, find the value of a+b+c?

Ans:       We know that,

  • \((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \)
  • \((x+y+z)^{2}=16+2(10) \)
  • \((x+y+z)^{2}=36\)
  • \((x+y+z)=\sqrt{36}\)
  • \((x+y+z)=\pm 6\)

Hence, value of required expression I; \((a+b+c)=\pm 8\)

Q5: If a+b+c=9 and ab+bc+ca=23, find value of \(a^{2}+b^{2}+c^{2}\)

Ans: We know that,

  • \((a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \)
  • \(9^{2}=a^{2}+b^{2}+c^{2}+2(23)\)
  • \(81=a^{2}+b^{2}+c^{2}+46\)
  • \(a^{2}+b^{2}+c^{2}=81-46\)
  • \(a^{2}+b^{2}+c^{2}=35\)

Hence, value of required expression \(a^{2}+b^{2}+c^{2}=35\)

Q6: Find the value of the equation : \(4x^{2}+y^{2}+25z^{2}+4xy-10yz-20zx\) when x=4,y=3,z=2

Ans:       \(4x^{2}+y^{2}+25z^{2}+4xy-10yz-20zx\)

  • \((2x)^{2}+y^{2}+(-5z)^{2}+2(2x)(y)+2(y)(-5z)+2(-5z)(2x)\)
  • \((2x+y-5z)^{2}\)
  • \((2(4)+3-5(2))^{2}\)
  • \((8+3-10)^{2}\)
  • \((1)^{2}\)
  • 1

Hence value of the equation is equals to 1

Q7: Simplify each of the following expressions:

 Q 7.1: \((x+y+z)^{2}+(x+\frac{y}{2}+\frac{2}{3})^{2}-(\frac{x}{2}+\frac{y}{3}+\frac{z}{4})^{2}\)

Ans:       Expanding, we get

\(=[x^{2}+y^{2}+z^{2}+2xy+2yz+2zx]+[x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}+2x\frac{y}{2}+2\frac{zx}{3}+\frac{yz}{3}]-[\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{z^{2}}{10}+\frac{xy}{3}+\frac{yz}{6}+\frac{xz}{4}]\)

\([∵(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx+x^{2}+\frac{y^{2}}{4}+\frac{z^{2}}{9}+2x\frac{y}{2}+\frac{xy}{3}+\frac{2zx}{3}-\frac{x^{2}}{4}-\frac{y^{2}}{9}-\frac{z^{2}}{10}-\frac{xy}{3}-\frac{yz}{6}-\frac{xz}{4}\)

Rearranging coefficients ,

\(=\frac{8x^{2}-x^{2}}{4}+\frac{36y^{2}+9y^{2}-4y^{2}}{36}+\frac{144z^{2}+16z^{2}-9z^{2}}{144}+\frac{6xy+3xy-xy}{3}+\frac{13yz}{6}+\frac{29xz}{12}\)

\(=\frac{7x^{2}}{4}+\frac{41y^{2}}{36}+\frac{151z^{2}}{144}+\frac{8xy}{3}+\frac{13yz}{6}+\frac{29zx}{12}\)

\((x+y+z)^{2}+(x+\frac{y}{2}+\frac{z}{3})^{2}-(\frac{x}{2}+\frac{y}{3}+\frac{z}{4})^{2}=\frac{7x^{2}}{4}+\frac{41y^{2}}{36}+\frac{151z^{2}}{144}+\frac{8xy}{3}+\frac{13yz}{6}+\frac{29zx}{12} \)

 Q 7.2: \((x+y-2z)^{2}-x^{2}-y^{2}-3z^{2}+4xy\)

                Ans:       \((x+y-2z)^{2}-x^{2}-y^{2}-3z^{2}+4xy\)

\(=[x^{2}+y^{2}+4z^{2}+2xy+2y(-2z)+2a(-2c)]-x^{2}-y^{2}-3z^{2}+4xy\)

\(=z^{2}+6xy-4yz-4zx\)

\((x+y-2z)^{2}-x^{2}-y^{2}-3z^{2}+4xy =z^{2}+6xy-4yz-4zx \)

 Q 7.3: \([x^{2}-x+1]^{2}-[x^{2}+x+1]^{2}\)

                Ans:       \([x^{2}-x+1]^{2}-[x^{2}+x+1]^{2}\)

\(=(x^{2})^{2}+(-x)^{2}+1^{2}+2(x^{2})(-x)+2(-x)(1)+2x^{2})-[(x^{2})^{2}+x^{2}+1+2x^{2}x+2x(1)+2x^{2}(1)]\)

\([∵ (x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2yz+2xz]\)

\(=x^{4}+y^{2}+1-2x^{3}-2x+2x^{2}-x^{2}-x^{4}-1-2x^{3}-2x-2x^{2}\)

\(=-4x^{3}-4x\)

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

Hence simplified equation =\([x^{2}-x+1]^{2}-[x^{2}+x+1]^{2}=-4x(x^{2}+1)\)

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