NCERT Solutions Class 9 Maths Ch 4: Algebraic Identities

ab < 0: a and b have opposite signs (one is positive, the other negative).

1. Expand the left side(a + b)² = a² + 2ab + b²
2. Put it in the inequalitya² + 2ab + b² < a² + b²
3. Subtract a² + b² from both sides2ab < 0 ⇒ ab < 0

The product of two numbers is negative only when one is positive and the other negative.

ab > 0: a and b have the same sign (both positive or both negative).

1. Use the expansion(a + b)² = a² + b² + 2ab
2. Compare(a + b)² > a² + b² ⇒ 2ab > 0 ⇒ ab > 0
3. ExampleTake a = 8 and b = 4: (8 + 4)² = 144, while a² + b² = 64 + 16 = 80, and 144 > 80. The figure shows the extra area 2ab = 2 × 32 = 64.

When ab = 0, that is, when a = 0 or b = 0.

1. Set them equala² + 2ab + b² = a² + b²
2. Cancel a² + b²2ab = 0 ⇒ ab = 0

So (a + b)² = a² + b² only when at least one of the two numbers is zero.

The term 2ab decides.

(a + b)² = a² + b² + 2ab, so the difference between the two is exactly 2ab.

We get a new identity: (a − b)² = a² − 2ab + b².

This is the identity for the square of a difference.

The three squares are a², b², c², and the six rectangles are ab (twice), ac (twice) and bc (twice).

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

35² = 1225, 65² = 4225, 85² = 7225, 105² = 11025. Every square ends in 25.

Use (a + b)² = a² + 2ab + b² with the number written as (tens + 5).

1. The patternThe square of a number ending in 5 always ends in 25. The digits before 25 are the tens part multiplied by the next number: 3 × 4 = 12 (1225), 6 × 7 = 42 (4225), 8 × 9 = 72 (7225), 10 × 11 = 110 (11025).
2. Why it works(10n + 5)² = 100n² + 100n + 25 = 100·n(n + 1) + 25.

(a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = 4a² + 4b² + 4c²

1. Read the top rowFour squares with sides a + b + c, a + b − c, a − b + c and a − b − c.
2. Read the bottom rowThree squares with sides 2a, 2b and 2c.
3. Both rows have the same total areaThe pieces are the same, only rearranged. So (a + b + c)² + (a + b − c)² + (a − b + c)² + (a − b − c)² = (2a)² + (2b)² + (2c)².
4. Check by expanding(a + b + c)² + (a + b − c)² = 2(a + b)² + 2c²(a − b + c)² + (a − b − c)² = 2(a − b)² + 2c²Sum = 2[(a + b)² + (a − b)²] + 4c² = 4a² + 4b² + 4c²

The PDF gave a product of the four terms instead of the sum of the squares. A product is not what the figure shows, so that answer has been replaced.

No, not for x² + 7x + 12. The split must be 3x + 4x, because 3 × 4 = 12.

In the textbook the tiles are x² + 7x + 12. To make a rectangle, the two parts of 7x must multiply to the constant 12.

The splitting 2x + 5x would work for x² + 7x + 10 = (x + 2)(x + 5), because 2 × 5 = 10.

(x + 2)(x + 3) = x² + 5x + 6

1. Lay out the tilesA rectangle with sides x + 3 and x + 2 needs: one x² tile, 3 + 2 = 5 x-tiles and 2 × 3 = 6 unit tiles.
2. Add the areasx² + 3x + 2x + 6 = x² + 5x + 6

x² + 11x + 30 = (x + 5)(x + 6)

1. Split 11xWe need two numbers with sum 11 and product 30: 5 and 6. So 11x = 5x + 6x.
2. Arrange the tilesPut the x² tile in the corner, 6 x-tiles along the top, 5 x-tiles down the side, and the 30 unit tiles (5 rows of 6) in the remaining corner.
3. Read the sidesThe rectangle has sides x + 6 and x + 5.

Both are correct. Reshma’s way is shorter.

1. JamesHe expands (a − b)² first and then multiplies by (a + b). This is correct but long.
2. ReshmaShe writes (a − b)² as (a − b)(a − b) and uses (a − b)(a + b) = a² − b². This is also correct and needs fewer steps.
3. The result(a − b)²(a + b) = (a − b)(a² − b²) = a³ − ab² − a²b + b³

New results by combining identities:

The blanks are (t + 8s)(t + (−6s)), and the simplified form is (t − 6s) ÷ (t + 8s).

1. Numerator36s² − 12st + t² = (6s − t)².
2. Denominatort² + 2ts − 48s²: we need two numbers with sum 2 and product −48. They are 8 and −6.t² + 2ts − 48s² = (t + 8s)(t − 6s)
3. SimplifySince (6s − t) = −(t − 6s), we have (6s − t)² = (t − 6s)².(t − 6s)²(t + 8s)(t − 6s) = t − 6st + 8s(This needs t ≠ 6s.)

(7x + 4y)² = 49x² + 56xy + 16y².

1. Identify a and bHere a = 7x and b = 4y.
2. Substitute in a² + 2ab + b²(7x)² + 2(7x)(4y) + (4y)²
3. Simplify49x² + 56xy + 16y²

(75x + 32y)² = 4925x² + 215xy + 94y².

1. Identify a and bHere a = 75x and b = 32y.
2. Substitute in a² + 2ab + b²(75x)² + 2(75x)(32y) + (32y)²
3. Simplify4925x² + 215xy + 94y²

(2.5p + 1.5q)² = 6.25p² + 7.5pq + 2.25q².

1. Identify a and bHere a = 2.5p and b = 1.5q.
2. Substitute in a² + 2ab + b²(2.5p)² + 2(2.5p)(1.5q) + (1.5q)²
3. Simplify6.25p² + 7.5pq + 2.25q²

(34s + 8t)² = 916s² + 12st + 64t².

1. Identify a and bHere a = 34s and b = 8t.
2. Substitute in a² + 2ab + b²(34s)² + 2(34s)(8t) + (8t)²
3. Simplify916s² + 12st + 64t²

(x + 12y)² = x² + xy + 14y².

1. Identify a and bHere a = x and b = 12y.
2. Substitute in a² + 2ab + b²x² + 2(x)(12y) + (12y)²
3. Simplifyx² + xy + 14y²

(1x + 1y)² = 1x² + 2xy + 1y².

1. Identify a and bHere a = 1x and b = 1y.
2. Substitute in a² + 2ab + b²(1x)² + 2(1x)(1y) + (1y)²
3. Simplify1x² + 2xy + 1y²

64² = 4096.

1. Split the number64 = 60 + 4.
2. Use the identity(60 + 4)² = 60² + 2(60)(4) + 4²
3. Calculate3600 + 480 + 16 = 4096

105² = 11025.

1. Split the number105 = 100 + 5.
2. Use the identity(100 + 5)² = 100² + 2(100)(5) + 5²
3. Calculate10000 + 1000 + 25 = 11025

205² = 42025.

1. Split the number205 = 200 + 5.
2. Use the identity(200 + 5)² = 200² + 2(200)(5) + 5²
3. Calculate40000 + 2000 + 25 = 42025

9x² + 24xy + 16y² = (3x + 4y)².

1. Recognise a perfect square9x² = (3x)², 16y² = (4y)² and 24xy = 2(3x)(4y).
2. Use a² + 2ab + b² = (a + b)²= (3x)² + 2(3x)(4y) + (4y)² = (3x + 4y)²

4s² + 20st + 25t² = (2s + 5t)².

1. Recognise a perfect square4s² = (2s)², 25t² = (5t)² and 20st = 2(2s)(5t).
2. Use the identity= (2s)² + 2(2s)(5t) + (5t)² = (2s + 5t)²

49x² + 28xy + 4y² = (7x + 2y)².

1. Recognise a perfect square49x² = (7x)², 4y² = (2y)² and 28xy = 2(7x)(2y).
2. Use the identity= (7x)² + 2(7x)(2y) + (2y)² = (7x + 2y)²

64p² + 323pq + 49q² = (8p + 23q)².

1. Recognise a perfect square64p² = (8p)², 49q² = (23q)² and 2(8p)(23q) = 323pq.
2. Use the identity= (8p)² + 2(8p)(23q) + (23q)² = (8p + 23q)²

3a² + 4ab + 43b² = 3(a + 23b)².

1. Take 3 common3a² + 4ab + 43b² = 3(a² + 43ab + 49b²)
2. Recognise a perfect square insidea² + 2(a)(23b) + (23b)² = (a + 23b)²
3. Put it together= 3(a + 23b)²

95s² + 6sv + 5v² = 15(3s + 5v)².

1. Take 15 common95s² + 6sv + 5v² = 15(9s² + 30sv + 25v²)
2. Recognise a perfect square9s² + 30sv + 25v² = (3s)² + 2(3s)(5v) + (5v)² = (3s + 5v)²
3. Put it together= 15(3s + 5v)²

79² = 6241.

1. Split the number79 = 80 − 1.
2. Use the identity(80 − 1)² = 80² − 2(80)(1) + 1²
3. Calculate6400 − 160 + 1 = 6241

193² = 37249.

1. Split the number193 = 200 − 7.
2. Use the identity(200 − 7)² = 200² − 2(200)(7) + 7²
3. Calculate40000 − 2800 + 49 = 37249

299² = 89401.

1. Split the number299 = 300 − 1.
2. Use the identity(300 − 1)² = 300² − 2(300)(1) + 1²
3. Calculate90000 − 600 + 1 = 89401

117² = 13689.

1. Split the number117 = 100 + 17.
2. Use the identity(100 + 17)² = 100² + 2(100)(17) + 17²
3. Calculate10000 + 3400 + 289 = 13689

78² = 6084.

1. Split the number78 = 80 − 2.
2. Use the identity(80 − 2)² = 80² − 2(80)(2) + 2²
3. Calculate6400 − 320 + 4 = 6084

198² = 39204.

1. Split the number198 = 200 − 2.
2. Use the identity(200 − 2)² = 200² − 2(200)(2) + 2²
3. Calculate40000 − 800 + 4 = 39204

214² = 45796.

1. Split the number214 = 200 + 14.
2. Use the identity(200 + 14)² = 200² + 2(200)(14) + 14²
3. Calculate40000 + 5600 + 196 = 45796

1104² = 1218816.

1. Split the number1104 = 1100 + 4.
2. Use the identity(1100 + 4)² = 1100² + 2(1100)(4) + 4²
3. Calculate1210000 + 8800 + 16 = 1218816

1120² = 1254400.

1. Split the number1120 = 1100 + 20.
2. Use the identity(1100 + 20)² = 1100² + 2(1100)(20) + 20²
3. Calculate1210000 + 44000 + 400 = 1254400

16y² − 24y + 9 = (4y − 3)².

1. Recognise a perfect square16y² = (4y)², 9 = 3² and 24y = 2(4y)(3).
2. Use a² − 2ab + b² = (a − b)²= (4y)² − 2(4y)(3) + 3² = (4y − 3)²

94s² + 6st + 4t² = (32s + 2t)².

1. Recognise a perfect square94s² = (32s)², 4t² = (2t)² and 2(32s)(2t) = 6st.
2. Use the identity= (32s)² + 2(32s)(2t) + (2t)² = (32s + 2t)²

The expression equals (m3 + k2 + 3n)².

1. Spot the squaresm²9 = (m3)², k²4 = (k2)² and 9n² = (3n)².
2. Check the cross terms2(m3)(k2) = mk3,   2(k2)(3n) = 3nk,   2(m3)(3n) = 2mn.
3. Use (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca= (m3 + k2 + 3n)²

p²16 − 2 + 16p² = (p4 − 4p)².

1. Recognise a perfect squarep²16 = (p4)², 16p² = (4p)² and 2(p4)(4p) = 2.
2. Use a² − 2ab + b² = (a − b)²= (p4)² − 2(p4)(4p) + (4p)² = (p4 − 4p)²

9a² + 4b² + c² − 12ab + 6ac − 4bc = (3a − 2b + c)².

1. Take the three terms as a = 3a, b = −2b, c = c(3a)² = 9a², (−2b)² = 4b², c² = c².
2. Check the cross terms2(3a)(−2b) = −12ab,   2(3a)(c) = 6ac,   2(−2b)(c) = −4bc
3. Use (a + b + c)²= (3a − 2b + c)²

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

1. Identifya = p, b = 3q, c = 7r.
2. Substitutep² + (3q)² + (7r)² + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)
3. Simplifyp² + 9q² + 49r² + 6pq + 42qr + 14pr

(3x − 2y + 4z)² = 9x² + 4y² + 16z² − 12xy − 16yz + 24xz.

1. Identifya = 3x, b = −2y, c = 4z.
2. Substitute(3x)² + (−2y)² + (4z)² + 2(3x)(−2y) + 2(−2y)(4z) + 2(4z)(3x)
3. Simplify9x² + 4y² + 16z² − 12xy − 16yz + 24xz

No, it is not an identity.

1. Expand the three squares(a + b − c)² = a² + b² + c² + 2ab − 2ac − 2bc(a − b + c)² = a² + b² + c² − 2ab + 2ac − 2bc(a − b − c)² = a² + b² + c² − 2ab − 2ac + 2bc
2. Add themLHS = 3a² + 3b² + 3c² − 2ab − 2ac − 2bc
3. CompareThe right side is 2a² + 2b² + 2c², which is different.
4. One example is enough to disprove itTake a = b = c = 1: LHS = 1² + 1² + (−1)² = 3, but RHS = 2 + 2 + 2 = 6.

Since LHS ≠ RHS, the statement is not an identity.

s² − 11s + 24 = (s − 3)(s − 8).

1. Find two numbersTheir sum is −11 and their product is 24. They are −3 and −8.
2. Write the factorss² − 11s + 24 = (s − 3)(s − 8)

The blank is 3x − 7.

1. Split the middle term3x² − 4x − 7 = 3x² − 7x + 3x − 7
2. Group= x(3x − 7) + 1(3x − 7) = (3x − 7)(x + 1)

Check: (3x − 7)(x + 1) = 3x² + 3x − 7x − 7 = 3x² − 4x − 7.

10x² − 11x − 6 = (2x − 3)(5x + 2). The blanks are 3 and 5x.

1. Look for the factorsThe first terms multiply to 10x² (2x and 5x) and the last terms to −6 (−3 and +2).
2. Check the middle term(2x − 3)(5x + 2) = 10x² + 4x − 15x − 6 = 10x² − 11x − 6

6x² + 7x + 2 = (3x + 2)(2x + 1).

1. Split 7xWe need two numbers with sum 7 and product 6 × 2 = 12. They are 3 and 4.
2. Split and group6x² + 3x + 4x + 2 = 3x(2x + 1) + 2(2x + 1) = (3x + 2)(2x + 1)

(41)² = 1681.

1. Identity chosen(a + b)² = a² + 2ab + b² with 41 = 40 + 1
2. Calculate(40 + 1)² = 1600 + 80 + 1 = 1681

(27)² = 729.

1. Identity chosen(a − b)² = a² − 2ab + b² with 27 = 30 − 3
2. Calculate(30 − 3)² = 900 − 180 + 9 = 729

23 × 17 = 391.

1. Identity chosen(a + b)(a − b) = a² − b² with 23 = 20 + 3 and 17 = 20 − 3
2. Calculate23 × 17 = (20 + 3)(20 − 3) = 20² − 3² = 400 − 9 = 391

(135)² = 18225.

1. Identity chosen(a + b)² = a² + 2ab + b² with 135 = 100 + 35
2. Calculate(100 + 35)² = 10000 + 7000 + 1225 = 18225

Shortcut for numbers ending in 5: 13 × 14 = 182, so 135² = 18225.

(97)² = 9409.

1. Identity chosen(a − b)² = a² − 2ab + b² with 97 = 100 − 3
2. Calculate(100 − 3)² = 10000 − 600 + 9 = 9409

18 × 29 = 522.

1. Identity chosen(x + a)(x + b) = x² + (a + b)x + ab with x = 20, a = −2, b = 9
2. Calculate18 × 29 = (20 − 2)(20 + 9) = 20² + (−2 + 9)(20) + (−2)(9) = 400 + 140 − 18 = 522

(34 × 43) = 1462.

1. Identity chosen(a + b)(a − b) = a² − b². Here 34 and 43 are not equal distances from a round number, so use their average and half-difference: a = (43 + 34) ÷ 2 = 77/2 and b = (43 − 34) ÷ 2 = 9/2
2. Calculate34 × 43 = (a − b)(a + b) = a² − b² = (772)² − (92)² = 59294 − 814 = 58484 = 1462

(205)² = 42025.

1. Identity chosen(a + b)² = a² + 2ab + b² with 205 = 200 + 5
2. Calculate(200 + 5)² = 40000 + 2000 + 25 = 42025

= (3a − b + 2c)².

1. Take a = 3a, b = −b, c = 2c(3a)² = 9a², (−b)² = b², (2c)² = 4c².
2. Check the cross terms2(3a)(−b) = −6ab,   2(3a)(2c) = 12ac,   2(−b)(2c) = −4bc
3. Use (a + b + c)²= (3a − b + 2c)²

= (4s − 5t)².

r² − r − 42 = (r − 7)(r + 6).

1. Find two numbersProduct −42 and sum −1. They are −7 and 6.
2. Write the factorsr² − r − 42 = (r − 7)(r + 6)

= (7g + h)².

= (8u − 11v − 2w)².

1. Take the terms 8u, −11v and −2w(8u)² = 64u², (−11v)² = 121v², (−2w)² = 4w².
2. Check the cross terms2(8u)(−11v) = −176uv,   2(8u)(−2w) = −32uw,   2(−11v)(−2w) = 44vw
3. Use (a + b + c)²= (8u − 11v − 2w)²

The expression simplifies to 3(p − 3q)(p + 2q)(p + 5q)(p − 2q). No more cancelling is possible.

1. NumeratorTake 3 common: 3p² − 3pq − 18q² = 3(p² − pq − 6q²) = 3(p − 3q)(p + 2q).
2. Denominatorp² + 3pq − 10q² = (p + 5q)(p − 2q), since 5 × (−2) = −10 and 5 + (−2) = 3.
3. Cancel common factorsNone of the factors (p − 3q), (p + 2q), (p + 5q), (p − 2q) match, so this is the simplest form.

The expression simplifies to n − m5.

1. Numeratorn³ − 3n²m + 3nm² − m³ = (n − m)³, using (a − b)³ = a³ − 3a²b + 3ab² − b³.
2. Denominator5m² − 10mn + 5n² = 5(m² − 2mn + n²) = 5(n − m)².
3. Cancel(n − m)³5(n − m)² = n − m5(This needs n ≠ m.)

The expression simplifies to w² + v² + x² + wv + vx − wxw − v + x.

1. Numerator: use a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)Take a = w, b = −v, c = x. Then a³ + b³ + c³ − 3abc = w³ − v³ + x³ + 3wvx, which is the numerator.w³ − v³ + x³ + 3wvx = (w − v + x)(w² + v² + x² + wv + vx − wx)
2. Denominatorw² + v² + x² − 2wv − 2vx + 2wx = (w − v + x)².
3. Cancel one factor (w − v + x)(w − v + x)(w² + v² + x² + wv + vx − wx)(w − v + x)² = w² + v² + x² + wv + vx − wxw − v + x

The PDF gave “w + x − v” as the answer, which is wrong: it used a wrong form of the identity. The correct second factor is w² + v² + x² + wv + vx − wx.

The expression simplifies to 5z − 2y5z + 2y.

1. Numerator4y² − 20yz + 25z² = (2y − 5z)² = (5z − 2y)².
2. Denominator25z² − 4y² = (5z)² − (2y)² = (5z − 2y)(5z + 2y).
3. Cancel (5z − 2y)(5z − 2y)²(5z − 2y)(5z + 2y) = 5z − 2y5z + 2y

The expression simplifies to 1.

1. Factorise each quadraticx² + x − 6 = (x + 3)(x − 2)x² − 7x + 12 = (x − 3)(x − 4)x² − 6x + 8 = (x − 2)(x − 4)x² − 9 = (x − 3)(x + 3)
2. Put them together and cancel(x + 3)(x − 2)(x − 3)(x − 4)(x − 2)(x − 4)(x − 3)(x + 3) = 1

Every factor on top cancels with one on the bottom.

The expression simplifies to (p + 2)(p² + 4)p − 2.

1. Numeratorp⁴ − 16 = (p²)² − 4² = (p² − 4)(p² + 4) = (p − 2)(p + 2)(p² + 4).
2. Denominatorp² − 4p + 4 = (p − 2)².
3. Cancel one (p − 2)(p − 2)(p + 2)(p² + 4)(p − 2)² = (p + 2)(p² + 4)p − 2

(−3x + 4)² = 9x² − 24x + 16.

1. Use (a + b)² with a = −3x, b = 4(−3x)² + 2(−3x)(4) + 4² = 9x² − 24x + 16

(2s + 7)(2s − 7) = 4s² − 49.

1. Use (a + b)(a − b) = a² − b² with a = 2s, b = 7(2s)² − 7² = 4s² − 49

(p² + 12)(p² − 12) = p⁴ − 14.

1. Use (a + b)(a − b) = a² − b² with a = p², b = 12(p²)² − (12)² = p⁴ − 14

(2n + 7)(2n − 7) = 4n² − 49.

1. Use (a + b)(a − b) = a² − b² with a = 2n, b = 7(2n)² − 7² = 4n² − 49

(s − 2t)(s² + 2st + 4t²) = s³ − 8t³.

1. Use (a − b)(a² + ab + b²) = a³ − b³ with a = s, b = 2ts³ − (2t)³ = s³ − 8t³

(12r − 4r)² = 14r² − 4 + 16r².

1. Use (a − b)² with a = 12r, b = 4r(12r)² − 2(12r)(4r) + (4r)² = 14r² − 4 + 16r²

(−3m + 4k − l)² = 9m² + 16k² + l² − 24mk − 8kl + 6lm.

1. Use (a + b + c)² with a = −3m, b = 4k, c = −la² = 9m², b² = 16k², c² = l².
2. Cross terms2ab = 2(−3m)(4k) = −24mk,   2bc = 2(4k)(−l) = −8kl,   2ca = 2(−l)(−3m) = 6lm
3. Add9m² + 16k² + l² − 24mk − 8kl + 6lm