EN हिं

NCERT Solutions for Class 9th Maths Chapter 4 Exercise Set 4.3 — More Identities

Book page 76–77 Updated on2026-09-08

Q1.
Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier. (i) 117²   (ii) 78²   (iii) 198²   (iv) 214²   (v) 1104²   (vi) 1120²
Answer

Pick the identity that keeps the pieces small: two-term identities when the number is close to a round number, the three-term identity when it is not.

(i) 117² — use (a + b + c)² with 100 + 10 + 7.

= 100² + 10² + 7² + 2(100)(10) + 2(10)(7) + 2(7)(100)
= 10000 + 100 + 49 + 2000 + 140 + 1400
= 13689

(ii) 78² — use (a − b)² with 80 − 2.

= 80² − 2 × 80 × 2 + 2² = 6400 − 320 + 4 = 6084

(iii) 198² — use (a − b)² with 200 − 2.

= 40000 − 800 + 4 = 39204

(iv) 214² — use (a + b + c)² with 200 + 10 + 4.

= 40000 + 100 + 16 + 2(200)(10) + 2(10)(4) + 2(4)(200)
= 40000 + 100 + 16 + 4000 + 80 + 1600 = 45796

(v) 1104² — use (a + b)² with 1100 + 4.

= 1100² + 2 × 1100 × 4 + 4² = 1210000 + 8800 + 16 = 1218816

(vi) 1120² — use (a + b)² with 1100 + 20.

= 1210000 + 2 × 1100 × 20 + 400 = 1210000 + 44000 + 400 = 1254400
Which identity, and why: a two-term split is enough when the number sits next to a round number — 78, 198, 1104, 1120 all do. But 117 and 214 are two digits away from anything round, so a two-term split would leave b = 17 or b = 14, and b² would be as hard as the original. Splitting into hundreds + tens + units makes every one of the six pieces a one-digit multiplication. The number of terms you need is the number of non-zero digits, so the three-term identity is the natural tool for a three-digit number.
Check it yourself: 1120² can also be done as (112 × 10)² = 112² × 100 = 12544 × 100. Same answer, and a good reminder that pulling out powers of ten first often saves work.
Q2.
Factor using suitable identities: (i) 16y² − 24y + 9   (ii) (9/4)s² + 6st + 4t²   (iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n²   (iv) p²/16 − 2 + 16/p²   (v) 9a² + 4b² + c² − 12ab + 6ac − 4bc
Answer

(i) 16y² − 24y + 9 — a square with a minus sign in the middle.

16y² = (4y)²,   9 = 3²,   2(4y)(3) = 24y ✓
= (4y − 3)²

(ii) (9/4)s² + 6st + 4t²

(9/4)s² = ((3/2)s)²,   4t² = (2t)²,   2((3/2)s)(2t) = 6st ✓
= ((3/2)s + 2t)²  — equivalently (1/4)(3s + 4t)²

(iii) m²/9 + mk/3 + k²/4 + 3nk + 2mn + 9n² — six terms, three of them squares, so try (a + b + c)².

m²/9 = (m/3)²,   k²/4 = (k/2)²,   9n² = (3n)²
Check the three cross terms with a = m/3, b = k/2, c = 3n:
2ab = 2(m/3)(k/2) = mk/3 ✓
2bc = 2(k/2)(3n) = 3kn ✓
2ca = 2(3n)(m/3) = 2mn ✓
= (m/3 + k/2 + 3n)²  — equivalently (1/36)(2m + 3k + 18n)²

(iv) p²/16 − 2 + 16/p² — the middle term is the clue: 2(p/4)(4/p) = 2.

p²/16 = (p/4)²,   16/p² = (4/p)²,   2(p/4)(4/p) = 2 ✓
= (p/4 − 4/p)²  — equivalently (p² − 16)²/(16p²), i.e. (p − 4)²(p + 4)²/(16p²)

(v) 9a² + 4b² + c² − 12ab + 6ac − 4bc — two of the three cross terms are negative, so one of the letters must carry a minus sign.

Try a′ = 3a, b′ = −2b, c′ = c:
2a′b′ = 2(3a)(−2b) = −12ab ✓
2b′c′ = 2(−2b)(c) = −4bc ✓
2c′a′ = 2(c)(3a) = 6ac ✓
= (3a − 2b + c)²
How to place the minus signs: the three cross terms of (a + b + c)² are 2ab, 2bc, 2ca. Changing the sign of exactly one letter flips exactly two of them — the two in which that letter appears. So a pattern of two minuses and one plus means one letter is negative; three minuses is impossible; and all pluses means none (or all) are negative. In (v) the negative terms are the ones containing b, so b is the letter to negate.
Q3.
Expand the following using the identity (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca: (i) (p + 3q + 7r)²   (ii) (3x − 2y + 4z)²
Answer

(i) (p + 3q + 7r)² — a = p, b = 3q, c = 7r.

= p² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
= p² + 9q² + 49r² + 6pq + 42qr + 14pr

(ii) (3x − 2y + 4z)² — take a = 3x, b = −2y, c = 4z and keep the signs inside the brackets.

= (3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)
= 9x² + 4y² + 16z² − 12xy − 16yz + 24zx
= 9x² + 4y² + 16z² − 12xy − 16yz + 24zx
Tip: Notice that the squared terms never lose their plus sign, however many minuses are inside — (−2y)² = +4y². Only the cross terms can turn negative.
Q4.
Is this an identity? (a + b − c)² + (a − b + c)² + (a − b − c)² = 2a² + 2b² + 2c²
Answer

No, it is not an identity. One substitution settles it, and the expansion shows exactly what has gone wrong.

Take a = b = c = 1.
Left side = (1 + 1 − 1)² + (1 − 1 + 1)² + (1 − 1 − 1)² = 1 + 1 + 1 = 3
Right side = 2 + 2 + 2 = 6
3 ≠ 6, so the equation fails at a perfectly ordinary point.

Expanding all three squares shows the true value of the left side:

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

That is not 2a² + 2b² + 2c², so the equation holds only for those special a, b, c satisfying a² + b² + c² = 2ab + 2bc + 2ca — an equation, not an identity.

The missing fourth term: the statement becomes true the moment you put (a + b + c)² back in. Adding it contributes a² + b² + c² + 2ab + 2bc + 2ca, and now every cross term cancels:
(a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = 4a² + 4b² + 4c²

The four sign patterns come in pairs that kill each cross term, and this is an identity — it is the very one drawn in Fig. 4.6 on page 78.

Tip: To disprove a claimed identity, one counter-example is enough and it should be the first thing you try. To prove one, you must expand.
Was this helpful? Report an error