NCERT Solutions for Class 9 Maths Chapter 4 Exploring Algebraic Identities (PDF) – 2027

Take four consecutive squares and compare the outer pair with the inner pair. The answer is always 4.

Now prove it. Four consecutive numbers can be written as (n − 1), n, (n + 1), (n + 2).

Every term containing n cancels, so the answer cannot depend on which four squares you picked.

Everything is decided by one term. Subtract:

So (a + b)² < a² + b² exactly when 2ab < 0, that is when

Check: a = 10, b = −2 gives (a + b)² = 8² = 64 while a² + b² = 100 + 4 = 104, and indeed 64 < 104.

Same difference, opposite sign.

Check both cases: a = 10, b = 2 gives 144 > 104. And a = −2, b = −3 gives (−5)² = 25 while a² + b² = 4 + 9 = 13, so 25 > 13.

For instance a = 7, b = 0: (7 + 0)² = 49 and 7² + 0² = 49.

(i) (7x + 4y)² — here a = 7x, b = 4y.

(ii) (7x/5 + 3y/2)² — a = 7x/5, b = 3y/2.

(iii) (2.5p + 1.5q)² — a = 2.5p, b = 1.5q.

(iv) (3s/4 + 8t)² — a = 3s/4, b = 8t.

(v) (x + 1/(2y))² — a = x, b = 1/(2y).

(vi) (1/x + 1/y)² — a = 1/x, b = 1/y.

Split each number so that one part is a round number and the other is small.

(i) 64² = (60 + 4)²

(ii) 105² = (100 + 5)²

(iii) 205² = (200 + 5)²

We get (a − b)² = a² − 2ab + b², which is also an identity.

Only the middle term changes sign, because it is the only term in which b appears an odd number of times.

(i) 9x² + 24xy + 16y²

(ii) 4s² + 20st + 25t²

(iii) 49x² + 28xy + 4y²

(iv) 64p² + (32/3)pq + (4/9)q²

*(v) 3a² + 4ab + (4/3)b² — yes, as the hint suggests, first take out a common factor, here 1/3:

*(vi) (9/5)s² + 6sv + 5v² — take out 1/5:

Each number is just below a round number, so subtract rather than add.

(i) 79² = (80 − 1)²

(ii) 193² = (200 − 7)²

(iii) 299² = (300 − 1)²

The side of the big square is split into a, b and c both horizontally and vertically, so the nine pieces have areas (row height) × (column width):

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.

(ii) 78² — use (a − b)² with 80 − 2.

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

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

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

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

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

(ii) (9/4)s² + 6st + 4t²

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

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

(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.

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

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

No, it is not an identity. One counter-example is enough:

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

This is not 2a² + 2b² + 2c², so the statement is only an equation, not an identity.

The figure is a cut-and-slide proof. Start with a square of side a and take a strip of width b off the bottom.

Every number here ends in 5, so Śhrīdharāchārya’s form a² = (a + b)(a − b) + b² with b = 5 is the right tool.

The pattern. Every answer ends in 25, and the digits in front are the tens digit multiplied by the next whole number:

The four squares of the top row together have the same area as the three squares of the bottom row:

Proof. Expand the four squares and add:

No — the split 2x + 5x does not work. If two of the x-tiles go to the right of the x²-tile and five go below it, then the unit tiles must fill a 2 by 5 corner, which holds only 10 units. We have 12, so two tiles are left over and the shape is not a rectangle.

There are only three ways to split 7x into two whole-number pieces. Check each:

Lay one x-tile and two unit tiles along the top edge, and one x-tile and three unit tiles down the left edge. Fill the rectangle:

Split 11x so that the unit tiles form a rectangle of area 30. We need a + b = 11 and ab = 30, so a = 5 and b = 6.

Look at where the numbers 7 and 12 came from, and where 13 and 42 came from:

Proof by the distributive property:

Count the tiles in Fig. 4.8. Two x-columns and three unit-columns across; three x-rows and one unit-row down.

Verify by the distributive property:

Verification by the distributive property:

(i) s² − 11s + 24 — two numbers with sum −11 and product 24 are −3 and −8.

(ii) (________)(x + 1) = 3x² − 4x − 7 — split −4x so that the given factor (x + 1) appears:

(iii) 10x² − 11x − 6 = (2x − ___)(___ + 2) — split −11x using two numbers with product 10 × (−6) = −60 and sum −11: they are −15 and 4.

(iv) 6x² + 7x + 2 — split 7x using two numbers with product 6 × 2 = 12 and sum 7: they are 3 and 4.

The two unequal products are worth writing out in full, since they use the third identity of Section 4.5.

(i) 9a² + b² + 4c² − 6ab + 12ac − 4bc — use (A + B + C)²; both negative cross terms contain b, so b takes the minus sign.

(ii) 16s² + 25t² − 40st

(iii) r² − r − 42 — two numbers with sum −1 and product −42 are −7 and 6.

(iv) 49g² + 14gh + h²

(v) 64u² + 121v² + 4w² − 176uv − 32uw + 44vw — both negative cross terms contain u, so v and w take the sign opposite to u: A = 8u, B = −11v, C = −2w.

Both are correct. They are two routes to the same expression, and both routes are legal. Reshma’s is the shorter one.

James’s route — expand the square first:

Reshma’s route — regroup first, so that a² − b² appears:

Prediction: x³ + y³. The chapter has just shown (x − y)(x² + xy + y²) = x³ − y³; replacing y by −y turns the left side into (x + y)(x² − xy + y²) and the right side into x³ + y³. Now confirm it by multiplying out.

Yes. The hint does almost all the work — the first bracket already contains x − y.

Yes. Here the trick of splitting into squares is not available (5 is odd), so multiply out directly and watch the cancellation.

Numerator (a perfect square):

Denominator: two terms with sum 2s and product −48s² are 8s and −6s.

So the blanks are filled by (t + 8s)(t − 6s), and now the fraction cancels:

(i) (3p² − 3pq − 18q²)/(p² + 3pq − 10q²)

No factor is common, so nothing cancels; the factorised form above is the answer.

(ii) (n³ − 3n²m + 3nm² − m³)/(5m² − 10mn + 5n²) — the numerator is a cube of a difference.

(iii) (w³ − v³ + x³ + 3wvx)/(w² + v² + x² − 2wv − 2vx + 2wx) — the numerator is the three-cubes identity with v negated.

(iv) (4y² − 20yz + 25z²)/(25z² − 4y²)

(v) [(x² + x − 6)(x² − 7x + 12)] / [(x² − 6x + 8)(x² − 9)]

(vi) (p⁴ − 16)/(p² − 4p + 4)

Put the x² square in a corner, 5 strips along its right side and 3 strips below it; the 15 unit squares fill the corner in 3 rows of 5.

(i) (−3x + 4)² — read it as (4 − 3x)² and use (a − b)².

(ii) (2s + 7)(2s − 7) — difference of two squares.

(iii) (p² + 1/2)(p² − 1/2)

(iv) (2n + 7)(2n − 7)

(v) (s − 2t)(s² + 2st + 4t²) — this is (x − y)(x² + xy + y²) with x = s, y = 2t.

(vi) (1/(2r) − 4r)² — a = 1/(2r), b = 4r, so 2ab = 2 × (1/(2r)) × 4r = 4.

(vii) (−3m + 4k − l)² — three-letter square with a = −3m, b = 4k, c = −l.

(viii) (x − y/3)³ — cube of a difference, a = x, b = y/3.

(ix) (7k/2 − 2m/3)³ — a = 7k/2, b = 2m/3.

(i) 17 × 21 — the two numbers are equally spaced about 19, so use (a + b)(a − b) = a² − b².

(ii) 104 × 96 — equally spaced about 100, so use a² − b².

(iii) 24 × 16 — equally spaced about 20.

(iv) 147³ = (150 − 3)³

(v) 199³ = (200 − 1)³

(vi) 127³ = (130 − 3)³

(vii) (−107)³ = −107³ = −(100 + 7)³

(viii) (−299)³ = −299³ = −(300 − 1)³

(i) 4y² + 1 + 1/(16y²) — the middle term 1 is the giveaway: 2 × 2y × 1/(4y) = 1.

(ii) 9m² − 1/(25n²) — difference of two squares.

(iii) 27b³ − 1/(64b³) — difference of two cubes, x = 3b, y = 1/(4b).

(iv) x² + 5x/6 + 1/6 — need a + b = 5/6 and ab = 1/6; take a = 1/2, b = 1/3.

(v) 27u³ − 27u²/5 + 9u/25 − 1/125 — four terms with alternating signs: a cube of a difference.

(vi) 64y³ + z³/125 — sum of two cubes, x = 4y, y′ = z/5.

(vii) p³ + 27q³ + r³ − 9pqr — the three-cubes identity, with 3q in place of the second letter.

(viii) 9m² − 12m + 4

(ix) 9x³ − (8/3)y³ + z³/3 + 6xyz — the coefficients are not cubes, so take out 1/3 first.

(x) 4x² + 9y² + 36z² + 12xz + 36yz + 24xy — the three squares point to 2x, 3y and 6z, and their cross terms are 2(2x)(3y) = 12xy, 2(3y)(6z) = 36yz, 2(6z)(2x) = 24zx. So the expression that factors is

(xi) 27u³ − 9u²/2 + u/4 − 1/216 — again a cube of a difference.

(i)

(ii) — take out the numerical factors first, then use the difference of cubes and the difference of squares.

(iii)

Area = length × breadth, so factorising the area into two factors gives a possible pair of dimensions.

(i) 25a² − 30ab + 9b²

Both sides are equal, so this rectangle is in fact a square.

(ii) 36s² − 49t²

Volume = length × breadth × height, so factor each expression into three factors.

(i) 6a² − 24b²

(ii) 3ps² − 15ps + 12p

The path is the region between two squares: the outer square (playground + path on both sides) and the playground itself.

Let the number be x. Then its reciprocal is 1/x, and x ≠ 0.

Split the middle term: two numbers with product 3 × 3 = 9 and sum −10 are −1 and −9.

Both answers are genuine, and they are reciprocals of each other.

Length = area ÷ width, so factor the area and see which factor is the given width.

If x − 2 is a factor, the expression must be zero at x = 2; likewise it must be zero at x = 1/2.

Subtract (2) from (1):

Use the three-cubes identity from Section 4.7:

The first bracket is given. For the second, first get a² + b² + c² from the square of the sum:

Now substitute both pieces:

Factor first.

So n³ − n is the product of three consecutive whole numbers. Now argue divisibility.

• Divisible by 2. Among any two consecutive integers one is even, so among three there is certainly an even one. Hence 2 divides the product.
• Divisible by 3. Any integer leaves remainder 0, 1 or 2 on division by 3. Among three consecutive integers, all three remainders occur exactly once, so one of them is a multiple of 3. Hence 3 divides the product.
• Therefore divisible by 6. The product is divisible by 2 and by 3, and 2 and 3 have no common factor other than 1, so their product 6 divides it.

(i) x³ + y³ − 12xy + 64, when x + y = −4

(ii) x³ − 8y³ − 36xy − 216, when x = 2y + 6