RD Sharma Solutions Ex-4.2, Algebraic Identities, Class 9, Maths

# RD Sharma Solutions Ex-4.2, Algebraic Identities, Class 9, Maths Notes | Study RD Sharma Solutions for Class 9 Mathematics - Class 9

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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): (a2+b2+c2)2

(vii): (ab+bc+ca)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 = a2+(2b)2+c2+2a(2b)+2ac+2(2b)c

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

?(a+2b+c)2 = a2+4b2+c2+4ab+2ac+4bc

Solution 1(ii):

We have,

(2a−3b−c)= [(2a)+(−3b)+(−c)]2

(2a)2+(−3b)2+(−c)2+2(2a)(−3b)+2(−3b)(−c)+2(2a)(−c)

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

4a2+9b2+c2−12ab+6bc−4ca

? (2a-3b-c)= 4x2+9y2+c2-12ab+6bc-4ca

Solution 1(iii):

We have,

(−3x+y+z)2 = [(−3x)2+y2+z2+2(−3x)y+2yz+2(−3x)z

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

9x2+y2+z2−6xy+2yz−6xz

(−3x+y+z)2 = 9x2+y2+z2−6xy+2xy−6xy

Solution 1(iv):

We have,

(m+2n−5p)2 = m2+(2n)2+(−5p)2+2m×2n+(2×2n×−5p)+2m×−5p

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

(m+2n−5p)2 = m2+4n2+25p2+4mn−20np−10pm

Solution 1(v):

We have,

(2+x−2y)2 = 22+x2+(−2y)2+2(2)(x)+2(x)(−2y)+2(2)(−2y)

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= 4+x2+4y2+4x−4xy−8y

(2+x−2y)2 = 4+x2+4y2+4x−4xy−8y

Solution 1(vi):

We have,

(a2+b2+c2)2 = (a2)2+(b2)2+(c2)2+2a2b2+2b2c2+2a2c2

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

(a2+b2+c2)2 = a4+b4+c4+2a2b2+2b2c2+2c2a2

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 = x2+y2+z2+2xy+2yz+2xz]

= a2b2+b2c2+c2a2+2(ac)b2+2(ab)(c)2+2(bc)(a)2

(ab+bc+ca)2 = a2b2+b2c2+c2a2+2acb2+2abc2+2bca2

Solution 1(viii):

We have,

Solution 1(ix):

We have,

Solution 1(x):

We have,

We have,

(x+2y+4z)2 = x2+(2y)2+(4z)2+2x×2y+2×2y×4z+2x×4z

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

(x+2y+4z)2 = x2+4y2+16z2+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 = x2+y2+z2+2xy+2yz+2xz]

(2x−y+z)2 = 4x2+y2+z2−4xy−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 = x2+y2+z2+2xy+2yz+2xz]

(−4x+6y+4z)2 = 4x2+9y2+4z2−12xy+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 = (a2+b2+c2+2ab+2bc+2ca)+(a2+(−b)2+c2−2ab−2bc+2ca)

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= 2a2+2b2+2c2+4ca

(a+b+c)2+(a−b+c)2 = 2a2+2b2+2c2+4ca

Q 2.2: (a+b+c)2−(a−b+c)2

Ans: We have,

(a+b+c)2−(a−b+c)= (a2+b2+c2+2ab+2bc+2ca)−(a2+(−b)2+c2−2ab−2bc+2ca)

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= a2+b2+c2+2ab+2bc+2ca−a2−b2−c2+2ab+2bc−2ca)

= 4ab+4bc

(a+b+c)2−(a−b+c)= 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)= a2+b2+c2+2ab+2bc+2ca+(a2+b2+(z)2−2bc−2ab+2ca)+(a2+b2+c2−2ca−2bc+2ab)

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= 3a2+3b2+3c2+2ab+2bc+2ca−2bc−2ab−2ca−2bc+2ab

= 3x2+3y2+3z2+2ab−2bc+2ca

(a+b+c)2+(a+b−c)2+(a−b+c)= 3a2+3b2+3c2+2ab−2bc+2ca

(a+b+c)2+(a+b−c)2+(a−b+c)= 3(a2+b2+c2)+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)= [2x2+p2+(−c)2+2(2x)p+2p(−c)+2(2x)(−c)]−[4x2+(−p)2+c2+2(2x)(−p)+2c(−p)+2(2x)c]

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

(2x+p−c)2−(2x−p+c)2 = [4x2+p2+c2+4xp−2pc−4xc]−[4x2+p2+c2−4xp−2pc+4xc]

Opening the bracket,

(2x+p−c)2−(2x−p+c)2 = 4x2+p2+c2+4xp−2pc−4cx−4x2−p2−c2+4xp+2pc−4cx]

(2x+p−c)2−(2x−p+c)= 8xp−8xc

= 8x(p−c)

Hence, (2x+p−c)2−(2x−p+c)= 8x(p−c)

Q 2.5: (x2+y2+(−z)2)−(x2−y2+z2)2

Ans: We have,

(x2+y2+(−z)2)2−(x2(−y)2+z2)2

=[x4+y4+(−z)4+2x2y2+2y2(−z)2+2x2(−z)2]−[x4+(−y)4+z4−2x2y2−2y2z22+2x2z2]

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= [x4+y4+(−z)4+2x2y2+2y2(−z)2+2x2(−z)2]−[x4+(−y)4+z4−2x2y2−2y2z2+2x2z2]

= 4x2y2–4z2x2

Hence, (x2+y2+(−z)2)2−(x2(−y)2+z2)= 4x2y2–4z2x2

Q3: If a+b+c = 0 and a2+b2+c2 = 16, find the value of ab+bc+ca:

Ans:  We know that,

[?(a+b+c)2 = a2+b2+c2+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 a2+b2+c2 = 16 and ab+bc+ca = 10, find the value of a+b+c?

Ans: We know that,

(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)

(x+y+z)2 = 16+2(10)

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

Q5: If a+b+c = 9 and ab+bc+ca = 23, find value of a2+b2+c2

Ans: We know that,

(a+b+c)2 = a2+b2+c2+2(ab+bc+ca)

92 = a2+b2+c2+2(23)

81 = a2+b2+c2+46

a2+b2+c2 = 81−46

a2+b2+c2 = 35

Hence, value of required expression a2+b2+c2 = 35

Q6: Find the value of the equation : 4x2+y2+25z2+4xy−10yz−20zx when x = 4,y = 3,z = 2

Ans: 4x2+y2+25z2+4xy−10yz−20zx

(2x)2+y2+(−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:

Ans: Expanding, we get

Rearranging coefficients ,

Q 7.2: (x+y−2z)2−x2−y2−3z2+4xy

Ans:  (x+y−2z)2−x2−y2−3z2+4xy

= [x2+y2+4z2+2xy+2y(−2z)+2a(−2c)]−x2−y2−3z2+4xy

= z2+6xy−4yz−4zx

(x+y−2z)2−x2−y2−3z2+4xy = z2+6xy−4yz−4zx

Q 7.3: [x2−x+1]2−[x2+x+1]2

Ans:  [x2−x+1]2−[x2+x+1]2

=(x2)2+(−x)2+12+2(x2)(−x)+2(−x)(1)+2x2)−[(x2)2+x2+1+2x2x+2x(1)+2x2(1)]

[?(x+y+z)2 = x2+y2+z2+2xy+2yz+2xz]

= x4+y2+1−2x3−2x+2x2−x2−x4−1−2x3−2x−2x2

= −4x3−4x

= −4x(x2+1)

Hence simplified equation = [x2−x+1]2−[x2+x+1]2 = −4x(x2+1)

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