Q.1. Expand:
(i) (a + 2b + 5c)^{2}
(ii) (2b − b + c)^{2}
(iii) (a − 2b − 3c)^{2}
Ans. (i) (a+2b+5c)^{2}
=(a)^{2} + (2b)^{2} +(5c)^{2}+2(a)(2b)+2(2b)(5c)+2(5c)(a) =a^{2}+4b^{2}+25c^{2}+4ab+20bc+10ac
(ii) (2a−b+c)^{2}=[(2a)+(−b)+(c)]^{2}
=(2a)^{2}+(−b)^{2}+(c)^{2}+2(2a)(−b)+2(−b)(c)+4(a)(c)
=4a^{2}+b^{2}+c^{2}−4ab−2bc+4ac
(iii) (a−2b−3c)^{2}=[a+(−2b)+(−3c)]^{2}
=(a)^{2}+(−2b)^{2}+(−3c)^{2}+2(a)(−2b)+2(−2b)(−3c)+2(a)(−3c)
=a^{2}+4b^{2}+9c^{2}−4ab+12bc−6ac
Q.2. Expand:
(i) (2a − 5b − 7c)^{2}
(ii) (−3a + 4b − 5c)^{2}
(iii) (12a−14a+2)^{2}
Ans. (i) (2a−5b−7c)^{2}
=[(2a)+(−5b)+(−7c)]^{2}
=(2a)^{2}+(−5b)^{2}+(−7c)^{2}+2(2a)(−5b)+2(−5b)(−7c)+2(2a)(−7c)
=4a^{2}+25b^{2}+49c^{2}−20ab+70bc−28ac
(ii) (−3a+4b−5c)^{2}=[(−3a)+(4b)+(−5c)]^{2}
=(−3a)^{2}+(4b)^{2}+(−5c)^{2}+2(−3a)(4b)+2(4b)(−5c)+2(−3a)(−5c)
=9a^{2}+16b^{2}+25c^{2}−24ab−40bc+30ac
(iii) (1/2a−1/4b+2)^{2}=[(a/2)+(−b/4)+(2)]^{2}
=(a^{2})^{2}+(−b^{4})^{2}+(2)^{2}+2(a/2)(−b/4)+2(−b/4)(2)+2(a/2)(2)
=a^{2}/4+b^{2}/16+4−ab/4−b+2a
Q.3. Factorize: 4x^{2} + 9y^{2} + 16z^{2} + 12xy − 24yz − 16xz.
Ans. We have: 4x^{2}+9y^{2}+16z^{2}+12xy−24yz−16xz
=(2x)^{2}+(3y)^{2}+(−4z)^{2}+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)
=[(2x)+(3y)+(−4z)]^{2}
=(2x+3y−4z)^{2}
Q. 4. Factorize: 9x^{2} + 16y^{2} + 4z^{2} − 24xy + 16yz − 12xz
Ans. We have: 9x^{2}+16y^{2}+4z^{2}−24xy+16yz−12xz
=(−3x)^{2}+(4y)^{2}+(2z)^{2}+2(−3x)(4y)+2(4y)(2z)+2(2z)(−3x)
=[(−3x)+(4y)+(2z)]^{2}
=(−3x+4y+2z)^{2}
Q.5. Factorize: 25x^{2} + 4y^{2} + 9z^{2} − 20xy − 12yz + 30xz.
Ans. We have: 25x^{2}+4y^{2}+9z^{2}−20xy−12yz+30xz
=(5x)^{2}+(−2y)^{2}+(3z)^{2}+2(5x)(−2y)+2(−2y)(3z)+2(3z)(5x)
=[(5x)+(−2y)+(3z)]^{2}
=(5x−2y+3z)^{2}
Q.6. 16x^{2} + 4y^{2} + 9z^{2} – 16xy – 12yz + 24xz
Ans. 16x^{2}+4y^{2}+9z^{2}−16xy−12yz+24xz
=(4x)^{2}+(−2y)^{2}+(3z)^{2}+2(4x)(−2y)+2(−2y)(3z)+2(3z)(4x)
=(4x−2y+3z)^{2} [using a^{2}+b^{2}+c^{2}+2ab+2bc+2ca=(a+b+c)^{2}]
Hence, 16x^{2} + 4y^{2} + 9z^{2} – 16xy – 12yz + 24xz = (4x−2y+3z)^{2}
Q.7. Evaluate
(i) (99)^{2}
(ii) (995)^{2}
(iii) (107)^{2}
Ans. (i) (99)^{2}=(100−1)^{2}
=[(100)+(−1)]^{2}
= (100)^{2}+2×(100)×(−1)+(−1)^{2}
=10000−200+1
=9801
(ii) (995)^{2}=(1000−5)^{2}
=[(1000)+(−5)]^{2}
=(1000)^{2}+2×(1000)×(−5)+(−5)^{2}
=1000000−10000+25
=990025
(iii) (107)^{2}=(100+7)^{2}
=(100)^{2}+2×(100)×(7)+(7)^{2}
=10000+1400+49
=11449
1 videos228 docs21 tests

1 videos228 docs21 tests


Explore Courses for Class 9 exam
