Matric Algebra, Coordinate Geometry and Commercial Maths

This chapter answers the Class IX-X mathematics items of the numerical papers in police recruitment exams: polynomials and their zeroes, linear equations in two variables, quadratic equations, arithmetic progressions, coordinate geometry, and the commercial strand of shares, dividends and instalments. Each topic is taught as the formula, one worked number and the slip that costs marks. Statistics is taught elsewhere and is not repeated. Every number below has been recomputed and checked by substitution.

Polynomials and their zeroes

A polynomial in x has only whole-number powers of x, such as 3x² − 5x + 7; its degree is the highest power (1 linear, 2 quadratic, 3 cubic). Expressions such as x + 1/x or √x + 2 are not polynomials. A zero of p(x) is a number k with p(k) = 0; geometrically it is where the graph meets the x-axis, so a quadratic has at most 2 zeroes and a cubic at most 3.

Relations between zeroes and coefficients
Polynomial	Sum of zeroes	Product of zeroes
ax² + bx + c	−b/a	c/a
ax³ + bx² + cx + d (cubic)	−b/a (pair sum c/a)	−d/a

Worked numbers. For 2x² − 8x + 6 the sum of zeroes is 8/2 = 4 and the product is 6/2 = 3 (the zeroes are 1 and 3). A quadratic with zeroes of sum 4 and product −5 is x² − 4x − 5. For x³ − 6x² + 11x − 6 the zeroes are 1, 2, 3: sum 6 = −(−6)/1, pair sum 2 + 3 + 6 = 11, product 6. If α and β are the zeroes of x² − 7x + 12, then α² + β² = (α + β)² − 2αβ = 49 − 24 = 25.

Remainder theorem: the remainder when p(x) is divided by (x − a) is p(a). Factor theorem: (x − a) is a factor exactly when p(a) = 0. Example: dividing x³ − 2x² + x − 5 by (x − 2) leaves p(2) = 8 − 8 + 2 − 5 = −3. For (x − 2) to be a factor of x² + kx − 10 we need 4 + 2k − 10 = 0, so k = 3.

🧠 Memory hook

Sign check for the sum: the zeroes' sum is the negative of the x-coefficient over the leading coefficient. For a quick test, the product c/a has the sign of c when a = 1.

Linear equations in two variables

A pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 represents two straight lines. The ratios decide how many solutions there are.

Condition	Lines	Solutions
a₁/a₂ ≠ b₁/b₂	Intersecting	Exactly one (consistent)
a₁/a₂ = b₁/b₂ ≠ c₁/c₂	Parallel	None (inconsistent)
a₁/a₂ = b₁/b₂ = c₁/c₂	Coincident	Infinitely many

Methods. Substitution (express one variable from one equation), elimination (make a coefficient equal and add or subtract) and cross-multiplication: x = (b₁c₂ − b₂c₁) / (a₁b₂ − a₂b₁), y = (c₁a₂ − c₂a₁) / (a₁b₂ − a₂b₁), for the form with +c on the left. Example: 2x + 3y = 12 and x − y = 1. From the second x = y + 1, so 2y + 2 + 3y = 12, 5y = 10, y = 2, x = 3; check: 6 + 6 = 12 and 3 − 2 = 1.

Finding k. For 2x + ky = 5 and 4x + 6y = 7 to have no solution, a₁/a₂ = b₁/b₂ requires 2/4 = k/6, so k = 3, and then c₁/c₂ = 5/7 ≠ 1/2, so the lines are truly parallel. Word problems. A two-digit number has digit sum 9 and the reversed number is 27 greater. With tens digit t and units digit u: t + u = 9 and (10u + t) − (10t + u) = 27, which gives u − t = 3, so u = 6, t = 3 and the number is 36 (reversed 63 = 36 + 27).

⚠️ Exam trap

Equal ratios of the coefficients do not by themselves mean infinitely many solutions: you must also compare the constants. a₁/a₂ = b₁/b₂ with c₁/c₂ different gives parallel lines and no solution.

Quadratic equations

The standard form is ax² + bx + c = 0 with a ≠ 0. Solve by factorising, or by the formula x = (−b ± √D) / 2a, where D = b² − 4ac is the discriminant. For x² − 7x + 12 = 0 the factors are (x − 3)(x − 4), giving 3 and 4. For 2x² − 5x + 3 = 0, D = 25 − 24 = 1 and x = (5 ± 1)/4, so x = 3/2 or 1.

Discriminant	Nature of roots	Example
D > 0	Two distinct real roots	x² − 5x + 6: D = 1
D = 0	Two equal real roots	2x² − 4x + 2: D = 0
D < 0	No real roots	x² + 4x + 5: D = −4

Finding k. For x² + kx + 9 = 0 to have equal roots, D = k² − 36 = 0, so k = 6 or −6. Word problem. The length of a rectangle is 3 m more than its breadth and the area is 70 m². With breadth x, x(x + 3) = 70, so x² + 3x − 70 = 0, which factorises as (x + 10)(x − 7) = 0. Length cannot be negative, so x = 7 m, length 10 m, perimeter 2(7 + 10) = 34 m. Always reject a root that is negative for a length, age or count.

⚠️ Exam trap

Two consecutive integers with product 132 are 11 and 12 or −12 and −11; the question decides whether negatives are allowed. Never state only one pair unless the stem says positive.

Arithmetic progressions

An arithmetic progression (AP) has a constant common difference d between consecutive terms. The n-th term is aₙ = a + (n − 1)d. The sum of the first n terms is Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l), where l is the last term. Three numbers a, b, c are in AP exactly when 2b = a + c.

Worked numbers. For 3, 7, 11, …: d = 4, a₁₀ = 3 + 9 × 4 = 39 and S₁₀ = 5 × (3 + 39) = 210. Which term of 5, 11, 17, … is 101? 5 + (n − 1) × 6 = 101 gives n − 1 = 16, so n = 17. Number of terms in 12, 16, …, 100: (100 − 12)/4 + 1 = 23. The multiples of 7 up to 100 are 7 to 98, 14 terms, with sum 14/2 × (7 + 98) = 735.

Two-condition problem. If the 3rd term is 12 and the 7th term is 24: a + 2d = 12 and a + 6d = 24, so 4d = 12, d = 3 and a = 6. Then S₁₀ = 5 × (2 × 6 + 9 × 3) = 5 × 39 = 195. Useful sums: 1 + 2 + … + n = n(n + 1)/2, so the first 100 natural numbers sum to 5050; the first n odd numbers sum to n².

⚠️ Exam trap

The n-th term uses (n − 1) common differences, not n. Writing a + nd is the single commonest AP error.

Coordinate geometry

A point is (x, y): x is the signed distance from the y-axis, y from the x-axis. The quadrants are I (+, +), II (−, +), III (−, −) and IV (+, −); a point on the x-axis has y = 0 and on the y-axis x = 0. The distance of (x, y) from the x-axis is |y|; its reflection in the x-axis is (x, −y).

Result	Formula
Distance	√[(x₂ − x₁)² + (y₂ − y₁)²]
Section formula (m : n, internal)	x = (mx₂ + nx₁)/(m + n), y = (my₂ + ny₁)/(m + n)
Midpoint	((x₁ + x₂)/2, (y₁ + y₂)/2)
Area of triangle	½ \|x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)\|
Centroid	((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3)

Worked numbers. Distance from (−2, 3) to (4, −5) = √(36 + 64) = 10. The point dividing (1, 2) and (7, 8) internally in the ratio 1 : 2 is ((1 × 7 + 2 × 1)/3, (1 × 8 + 2 × 2)/3) = (3, 4). The midpoint of (2, −3) and (8, 5) is (5, 1). The triangle (1, 1), (5, 1), (1, 4) has area ½ |1(1 − 4) + 5(4 − 1) + 1(1 − 1)| = ½ × 12 = 6 square units (base 4, height 3). The centroid of (0, 0), (6, 0), (0, 9) is (2, 3).

Collinearity: three points lie on one line exactly when the triangle area is zero. For (1, 2), (2, 4), (k, 6): ½ |1(4 − 6) + 2(6 − 2) + k(2 − 4)| = ½ |6 − 2k| = 0, so k = 3. The slope of the line through two points is (y₂ − y₁)/(x₂ − x₁); through (1, 2) and (3, 8) it is 3.

⚠️ Exam trap

In the section formula the ratio m : n is paired crosswise: m multiplies the second point's coordinates, n the first's. Writing it the other way gives the point the same distance from the wrong end.

Commercial maths: shares, dividends and instalments

A company's capital is divided into shares of a fixed face value (FV, also called nominal value). The market value (MV) is the price at which a share trades: above FV is at a premium, below FV at a discount, equal to FV is at par. The dividend is declared as a percentage of the face value, never of the market value. Income = number of shares × FV × rate; yield = dividend per share / MV × 100.

Worked numbers. Buy 50 shares of FV ₹100 at MV ₹120 with a 9% dividend. Investment = 50 × 120 = ₹6,000. Dividend per share = ₹9, so annual income = 50 × 9 = ₹450 and yield = 450/6000 × 100 = 7.5%. If ₹9,000 is put into ₹100 shares quoted at ₹90 (10% discount), the number of shares is 9000/90 = 100; an 8% dividend then gives 100 × 8 = ₹800.

Instalments. In a hire-purchase scheme the buyer pays part now and the rest in instalments; the extra over the cash price is interest. Cash price ₹10,000, paid as ₹2,000 down plus 8 monthly instalments of ₹1,100: total paid = 2000 + 8800 = ₹10,800, so ₹800 is the interest charged. For a loan repaid in equal yearly instalments with compound interest, a loan of ₹2,100 at 10% a year in two equal instalments needs x with x/1.1 + x/1.21 = 2100, so x = ₹1,210. Check: after year 1 the debt is 2310; pay 1210 leaves 1100; after year 2 it is 1210; pay 1210 clears it.

⚠️ Exam trap

Dividend is on face value, yield is on market value. A 9% dividend share bought at ₹120 does not yield 9%; it yields 9/120 × 100 = 7.5% when FV is ₹100.

Key takeaways

A polynomial has only whole-number powers. For ax² + bx + c the zeroes sum to −b/a and multiply to c/a; remainder of p(x) ÷ (x − a) is p(a), and zero remainder means a factor.
A pair of linear equations has one, none or infinitely many solutions according as a₁/a₂ ≠ b₁/b₂, or = b₁/b₂ ≠ c₁/c₂, or all three ratios equal.
Quadratic roots: x = (−b ± √D)/2a with D = b² − 4ac; D > 0 two real roots, D = 0 equal, D < 0 none. Reject negative roots for lengths and ages.
AP: aₙ = a + (n − 1)d and Sₙ = n/2 [2a + (n − 1)d]. Sum of first n natural numbers is n(n + 1)/2; of first n odd numbers is n².
Coordinates: distance, section formula (crosswise m : n), midpoint, centroid, and triangle area (zero means collinear).
Shares: dividend is a percentage of face value; yield is dividend over market value; instalment interest equals total paid minus cash price.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which of the following is NOT a polynomial in x?
1. 3x² − 5x + 7
2. x + 1/x
3. x³ + 4
4. 7
Show answer
Answer: B — x + 1/x
A polynomial allows only non-negative whole-number powers of x. The term 1/x is x to the power −1, so x + 1/x is not a polynomial. A constant such as 7 is a polynomial of degree 0, which is the tempting wrong choice.
The remainder when x³ − 2x² + x − 5 is divided by (x − 2) is
1. 3
2. −5
3. −3
4. 0
Show answer
Answer: C — −3
By the remainder theorem the remainder is p(2) = 8 − 8 + 2 − 5 = −3. A zero remainder would mean (x − 2) is a factor, which p(2) = −3 rules out; −5 is just p(0).
If α and β are the zeroes of x² − 7x + 12, the value of α² + β² is
1. 49
2. 37
3. 13
4. 25
Show answer
Answer: D — 25
The sum is 7 and the product is 12, so α² + β² = 49 − 2 × 12 = 25 (the zeroes are 3 and 4, and 9 + 16 = 25). The value 49 is (α + β)² and forgets to subtract 2αβ.
For which value of k does the pair 2x + ky = 5 and 4x + 6y = 7 have no solution?
1. 3
2. 12
3. 5/7
4. 6
Show answer
Answer: A — 3
No solution needs parallel lines: 2/4 = k/6 gives k = 3, and 5/7 differs from 1/2, so the lines do not coincide. For any other k the ratios of x and y coefficients differ and there is exactly one solution.
The digits of a two-digit number add up to 9, and on reversing the digits the number increases by 27. The original number is
1. 63
2. 45
3. 36
4. 27
Show answer
Answer: C — 36
With tens digit t and units digit u, t + u = 9 and (10u + t) − (10t + u) = 9(u − t) = 27, so u − t = 3, giving u = 6 and t = 3. The number is 36 and its reverse 63 is 27 more. The choice 63 is the reversed number, not the original.
The roots of x² − 7x + 12 = 0 are
1. −3 and −4
2. 3 and 4
3. 2 and 6
4. 1 and 12
Show answer
Answer: B — 3 and 4
Factorising gives (x − 3)(x − 4) = 0, so the roots are 3 and 4, which sum to 7 and multiply to 12. The pair −3 and −4 has the right product but the wrong sum (−7).
Judge these statements. P: x² + 4x + 5 = 0 has no real roots. Q: 2x² − 4x + 2 = 0 has two equal real roots. R: x² − 5x + 6 = 0 has roots 2 and 3. S: x² + 1 = 0 has two distinct real roots. Which are correct?
1. P, Q and S only
2. Q, R and S only
3. P, Q and R only
4. P and R only
Show answer
Answer: C — P, Q and R only
The discriminant of P is 16 − 20 = −4, so no real roots; Q gives 16 − 16 = 0, equal roots; R factorises as (x − 2)(x − 3). S has D = −4, so no real roots, which makes it false. The set P and R only misses that Q is also true.
For what values of k does x² + kx + 9 = 0 have two equal real roots?
1. k = 6 only
2. k = 3 or k = −3
3. k = 36 or k = −36
4. k = 6 or k = −6
Show answer
Answer: D — k = 6 or k = −6
Equal roots need D = k² − 4 × 1 × 9 = 0, so k² = 36 and k = ±6. The answer k = 6 only drops the negative root of k² = 36, and 36 is the value of k², not of k.
The length of a rectangular field is 3 m more than its breadth and its area is 70 m². Its perimeter is
1. 34 m
2. 17 m
3. 26 m
4. 40 m
Show answer
Answer: A — 34 m
With breadth x, x(x + 3) = 70 gives x² + 3x − 70 = (x + 10)(x − 7) = 0, so x = 7 and length 10. The perimeter is 2 × (7 + 10) = 34 m. The value 17 is only half of the perimeter, the sum of length and breadth.
The 10th term of the arithmetic progression 3, 7, 11, 15, … is
1. 43
2. 39
3. 40
4. 37
Show answer
Answer: B — 39
Here a = 3 and d = 4, so a₁₀ = 3 + 9 × 4 = 39. The value 43 is the 11th term, obtained by using n instead of n − 1 differences.
Which term of the arithmetic progression 5, 11, 17, 23, … is 101?
1. 16th
2. 18th
3. 17th
4. 15th
Show answer
Answer: C — 17th
5 + (n − 1) × 6 = 101 gives (n − 1) = 16, so n = 17. The number 16 is the value of n − 1, which is the commonest slip, and the 16th term is 95, not 101.
The sum of the first 10 terms of the arithmetic progression 3, 7, 11, … is
1. 195
2. 420
3. 390
4. 210
Show answer
Answer: D — 210
S₁₀ = 10/2 × [2 × 3 + 9 × 4] = 5 × 42 = 210. Equivalently 5 × (first + last) = 5 × (3 + 39). The value 420 forgets the factor ½ in n/2.
In an arithmetic progression the 3rd term is 12 and the 7th term is 24. The sum of its first 10 terms is
1. 210
2. 195
3. 180
4. 165
Show answer
Answer: B — 195
a + 2d = 12 and a + 6d = 24 give d = 3 and a = 6. Then S₁₀ = 5 × (2 × 6 + 9 × 3) = 195. The tempting 210 is the sum of a different progression (3, 7, 11, …), not this one.
The distance between the points (−2, 3) and (4, −5) is
1. 10 units
2. 14 units
3. 2 units
4. √28 units
Show answer
Answer: A — 10 units
Distance = √[(4 + 2)² + (−5 − 3)²] = √(36 + 64) = √100 = 10. The value 14 adds the differences (6 + 8) instead of squaring and taking the root.
The point that divides the line segment joining (1, 2) and (7, 8) internally in the ratio 1 : 2 (measured from the first point) is
1. (5, 6)
2. (3, 4)
3. (4, 5)
4. (2, 3)
Show answer
Answer: B — (3, 4)
x = (1 × 7 + 2 × 1)/3 = 3 and y = (1 × 8 + 2 × 2)/3 = 4, so the point is (3, 4), one-third of the way from (1, 2). The point (5, 6) is the 2 : 1 division, from using the ratio the wrong way round.
The points (1, 2), (2, 4) and (k, 6) lie on one straight line. The value of k is
1. 2
2. 4
3. 3
4. 6
Show answer
Answer: C — 3
Collinear points give zero area: ½ |1(4 − 6) + 2(6 − 2) + k(2 − 4)| = ½ |6 − 2k| = 0, so k = 3. Equivalently the slope from (1, 2) to (2, 4) is 2, and (3, 6) keeps that slope.
A person buys 50 shares of face value ₹100 at a market value of ₹120 each, and the company declares a 9% dividend. The percentage return on the money invested is
1. 9%
2. 10.8%
3. 6%
4. 7.5%
Show answer
Answer: D — 7.5%
Dividend per share is 9% of ₹100 = ₹9; return = 9/120 × 100 = 7.5% on the ₹120 paid. The answer 9% is the dividend rate on face value, not the return on market value; 10.8% wrongly applies 9% to the market price.
A loan of ₹2,100 at 10% compound interest per year is to be cleared in two equal annual instalments, the first paid one year after the loan. Each instalment is
1. ₹1,210
2. ₹1,155
3. ₹1,100
4. ₹1,260
Show answer
Answer: A — ₹1,210
The instalment x satisfies x/1.1 + x/1.21 = 2100, so x × 2.1/1.21 = 2100 and x = 1210. Check: 2100 grows to 2310, minus 1210 leaves 1100, which grows to 1210 and is cleared. The figure 1,155 is half of 2310, which ignores the interest on the unpaid second half.