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)2 = [(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)2 = 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)2 = (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)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 = 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)2 = 3a2+3b2+3c2+2ab−2bc+2ca
(a+b+c)2+(a+b−c)2+(a−b+c)2 = 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)2 = [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)2 = 8xp−8xc
= 8x(p−c)
Hence, (2x+p−c)2−(2x−p+c)2 = 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]
Taking the negative sign inside,
= [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)2 = 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)
1. What are algebraic identities? |
2. What are some common algebraic identities used in Class 9 Maths? |
3. How can algebraic identities be used to simplify expressions? |
4. Are algebraic identities only applicable in algebra? |
5. Can algebraic identities be used to solve real-life problems? |
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