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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): (a^{2}+b^{2}+c^{2})^{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} = 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
? (2a3bc)^{2 }= 4x^{2}+9y^{2}+c^{2}12ab+6bc4ca
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×2n+(2×2n×−5p)+2m×−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}+4y2+4x−4xy−8y
Solution 1(vi):
We have,
(a^{2}+b2+c2)^{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,
Solution 1(ix):
We have,
Solution 1(x):
We have,
We have,
(x+2y+4z)^{2} = x^{2}+(2y)^{2}+(4z)^{2}+2x×2y+2×2y×4z+2x×4z
[?(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}+c2+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}z2^{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}+y4+(−z)^{4}+2x^{2}y^{2}+2y^{2}(−z)^{2}+^{2}x^{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)
Hence, value of required expression I; (a+b+c)=±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:
Ans: Expanding, we get
Rearranging coefficients ,
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}+12+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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