Chapter 8 — Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0. This is the standard form. The condition a ≠ 0 is essential: without the x² term there is no quadratic. The number a is the coefficient of x², b the coefficient of x and c the constant term; b or c may be zero.

Examples in standard form: 2x² − 3x + 1 = 0 (a = 2, b = −3, c = 1); x² − 9 = 0 (a = 1, b = 0, c = −9); 3x² + 5x = 0 (a = 3, b = 5, c = 0); x² = 0 (a = 1, b = c = 0).

Deciding whether an equation is quadratic. Expand and simplify, bring everything to one side, and look for the highest power. (x + 1)² = 2(x − 3): expand to x² + 2x + 1 = 2x − 6, so x² + 7 = 0, quadratic. x² − 2x = (−2)(3 − x): x² − 2x = −6 + 2x, so x² − 4x + 6 = 0, quadratic. (x − 2)(x + 1) = (x − 1)(x + 3): x² − x − 2 = x² + 2x − 3, so −3x + 1 = 0, which is linear, not quadratic, because the x² terms cancel. (x + 2)³ = 2x(x² − 1): x³ + 6x² + 12x + 8 = 2x³ − 2x, so −x³ + 6x² + 14x + 8 = 0, cubic, not quadratic. x³ − 4x² − x + 1 = (x − 2)³: x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8, so 2x² − 13x + 9 = 0, quadratic, because the cubes cancel.

Roots. A real number α is a root (or solution) of ax² + bx + c = 0 if aα² + bα + c = 0. The roots of the equation are exactly the zeroes of the polynomial ax² + bx + c. So a quadratic equation has at most two real roots. To check whether 2 is a root of x² − 5x + 6 = 0: 4 − 10 + 6 = 0, yes. Is −1 a root of 2x² + x − 1 = 0? 2 − 1 − 1 = 0, yes. Is 3 a root of x² + 2x − 3 = 0? 9 + 6 − 3 = 12 ≠ 0, no.

From words to equations. A plot's length is one more than twice its breadth and its area is 528 square metres. Let breadth be x; length is 2x + 1; area x(2x + 1) = 528, so 2x² + x − 528 = 0. Two consecutive positive integers have product 306: x(x + 1) = 306, so x² + x − 306 = 0. Rohan's mother is 26 years older than him and the product of their ages three years from now will be 360: (x + 3)(x + 29) = 360, so x² + 32x − 273 = 0. A train travels 480 km at a uniform speed; if the speed were 8 km/h less it would take 3 hours more: 480/(x − 8) − 480/x = 3, which clears to x² − 8x − 1280 = 0. Forming the equation is the first mark in every word problem, and writing it in standard form the second.

If a product of two factors is zero, at least one factor must be zero. This zero product rule is the basis of solving quadratics by factorisation: write ax² + bx + c as a product of two linear factors, set each factor equal to zero, and solve the two linear equations.

Splitting the middle term. To factorise ax² + bx + c, find two numbers p and q with p × q = a × c and p + q = b, rewrite bx as px + qx, and factor by grouping.

Example 1. x² − 3x − 10 = 0. a × c = −10, b = −3: numbers −5 and 2. x² − 5x + 2x − 10 = x(x − 5) + 2(x − 5) = (x − 5)(x + 2) = 0. So x = 5 or x = −2.

Example 2. 2x² + x − 6 = 0. a × c = −12, b = 1: numbers 4 and −3. 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (x + 2)(2x − 3) = 0. So x = −2 or x = 3/2.

Example 3. √2 x² + 7x + 5√2 = 0. a × c = √2 × 5√2 = 10, b = 7: numbers 5 and 2. √2 x² + 5x + 2x + 5√2 = x(√2 x + 5) + √2(√2 x + 5) = (√2 x + 5)(x + √2) = 0. So x = −5/√2 or x = −√2.

Example 4. 2x² − x + 1/8 = 0. Multiply by 8: 16x² − 8x + 1 = 0. a × c = 16, b = −8: numbers −4 and −4. 16x² − 4x − 4x + 1 = 4x(4x − 1) − 1(4x − 1) = (4x − 1)² = 0. So x = 1/4, a repeated root.

Example 5. 100x² − 20x + 1 = 0: (10x − 1)² = 0, x = 1/10 (repeated).

Example 6. 3x² − 2√6 x + 2 = 0. a × c = 6, b = −2√6: numbers −√6 and −√6. 3x² − √6 x − √6 x + 2 = √3 x(√3 x − √2) − √2(√3 x − √2) = (√3 x − √2)² = 0. So x = √2/√3 = √(2/3), repeated.

Example 7. x² − 45x + 324 = 0 (from a word problem about two numbers whose sum is 45 and product 324). Numbers −9 and −36: (x − 9)(x − 36) = 0, x = 9 or 36.

Special forms. Difference of squares: x² − 16 = (x − 4)(x + 4), roots ±4. Common factor: 3x² − 12x = 3x(x − 4), roots 0 and 4. Perfect squares: x² + 6x + 9 = (x + 3)², root −3 repeated.

When factorisation is hard. If no integer pair p, q can be found, the roots are irrational or non-real, and completing the square or the formula is needed. For example x² − 4x + 1 = 0 has no integer split (product 1, sum −4 is impossible with integers), and its roots are 2 ± √3.

After solving, substitute each root back. Roots like 3/2 come from the factor 2x − 3, not from the number 3; this is the most frequent slip when a ≠ 1.

Not every quadratic factorises neatly. Completing the square is a method that always works: it rewrites the equation so that x appears only inside a squared bracket, then takes square roots.

The idea. x² + 2kx becomes a perfect square when k² is added: x² + 2kx + k² = (x + k)². So for x² + bx, add (b/2)², half the coefficient of x, squared.

Steps for ax² + bx + c = 0. (1) If a ≠ 1, divide the whole equation by a. (2) Move the constant to the right side. (3) Add the square of half the coefficient of x to both sides. (4) Write the left side as a perfect square. (5) Take square roots of both sides, remembering ±. (6) Solve for x.

Example 1. x² + 4x − 5 = 0. Move the constant: x² + 4x = 5. Half of 4 is 2, square 4: x² + 4x + 4 = 9. So (x + 2)² = 9, x + 2 = ±3, x = 1 or x = −5.

Example 2. 2x² − 7x + 3 = 0. Divide by 2: x² − (7/2)x + 3/2 = 0. Move: x² − (7/2)x = −3/2. Half of 7/2 is 7/4, square 49/16: x² − (7/2)x + 49/16 = −3/2 + 49/16 = −24/16 + 49/16 = 25/16. So (x − 7/4)² = 25/16, x − 7/4 = ±5/4, x = 12/4 = 3 or x = 2/4 = 1/2.

Example 3. x² − 4x + 1 = 0. x² − 4x = −1. Add 4: (x − 2)² = 3, x − 2 = ±√3, x = 2 ± √3. Here factorisation would have failed, but the square gives exact irrational roots.

Example 4. 4x² + 4√3 x + 3 = 0. Divide by 4: x² + √3 x + 3/4 = 0. x² + √3 x = −3/4. Add (√3/2)² = 3/4: (x + √3/2)² = 0. So x = −√3/2, a repeated root.

Example 5. 2x² + x + 4 = 0. Divide by 2: x² + x/2 + 2 = 0. x² + x/2 = −2. Add 1/16: (x + 1/4)² = −2 + 1/16 = −31/16. A square of a real number cannot be negative, so there is no real root. Completing the square reveals this directly.

Example 6. 5x² − 6x − 2 = 0. Divide by 5: x² − (6/5)x = 2/5. Add (3/5)² = 9/25: (x − 3/5)² = 10/25 + 9/25 = 19/25. x = 3/5 ± √19/5 = (3 ± √19)/5.

An alternative without fractions. Multiply ax² + bx + c = 0 by 4a: 4a²x² + 4abx + 4ac = 0, so (2ax + b)² = b² − 4ac. For 2x² − 7x + 3 = 0: (4x − 7)² = 49 − 24 = 25, 4x − 7 = ±5, x = 3 or 1/2. This variant keeps integers and is a step away from the formula.

Completing the square is examined as a method in its own right; when the question says by completing the square, the perfect square step must be shown explicitly, and the ± at the square root must not be forgotten.

Completing the square on the general equation gives a formula that solves every quadratic at once.

Derivation. Start with ax² + bx + c = 0, a ≠ 0. Divide by a: x² + (b/a)x + c/a = 0. Move the constant: x² + (b/a)x = −c/a. Add (b/2a)² = b²/4a² to both sides: (x + b/2a)² = b²/4a² − c/a = (b² − 4ac)/4a². If b² − 4ac ≥ 0, take square roots: x + b/2a = ±√(b² − 4ac)/2a. Hence

x = [−b ± √(b² − 4ac)] / 2a.

This is the quadratic formula. The two signs give the two roots. It is due to the Indian mathematician Sridharacharya (around 1025 CE) and is sometimes called Sridharacharya's formula.

Example 1. 2x² − 7x + 3 = 0: a = 2, b = −7, c = 3. b² − 4ac = 49 − 24 = 25. x = [7 ± 5]/4 = 3 or 1/2.

Example 2. x² + 4x + 5 = 0: b² − 4ac = 16 − 20 = −4 < 0. No real roots.

Example 3. 3x² − 5x + 2 = 0: b² − 4ac = 25 − 24 = 1. x = [5 ± 1]/6 = 1 or 2/3.

Example 4. x² − 2x − 4 = 0: b² − 4ac = 4 + 16 = 20. x = [2 ± √20]/2 = [2 ± 2√5]/2 = 1 ± √5. Simplify surds before dividing.

Example 5. 4x² + 4√3 x + 3 = 0: b² − 4ac = 48 − 48 = 0. x = −4√3/8 = −√3/2, one repeated root.

Example 6. x − 1/x = 3, x ≠ 0. Multiply by x: x² − 3x − 1 = 0. b² − 4ac = 9 + 4 = 13. x = (3 ± √13)/2.

Example 7. 1/(x + 4) − 1/(x − 7) = 11/30, x ≠ −4, 7. Combine: [(x − 7) − (x + 4)]/[(x + 4)(x − 7)] = 11/30, so −11/(x² − 3x − 28) = 11/30, giving x² − 3x − 28 = −30, x² − 3x + 2 = 0. b² − 4ac = 9 − 8 = 1. x = (3 ± 1)/2 = 2 or 1. Both are allowed (neither is −4 or 7).

Using the formula well. Write a, b, c with their signs before substituting. Compute b² − 4ac separately and note its sign. Simplify the square root (√20 = 2√5, √72 = 6√2). Divide every term of the numerator by 2a. Check one root by substitution.

Choosing a method. Factorise when integer splits are obvious. Complete the square when the question demands it or when a = 1 and b is even. Use the formula for everything else, especially when b² − 4ac is not a perfect square. All three give the same roots; the formula is the universal fallback.

In the quadratic formula the quantity under the square root, b² − 4ac, is called the discriminant and is written D or Δ. It discriminates, that is, distinguishes, between the possible kinds of roots without the roots being found.

Three cases. If D > 0, the square root is a positive real number and the ± gives two different values: two distinct real roots. If D = 0, the square root is zero and both signs give the same value −b/2a: two equal real roots (one repeated root). If D < 0, there is no real square root: no real roots.

Geometrically these correspond to the parabola y = ax² + bx + c cutting the x-axis at two points, touching it at one point, or missing it entirely.

Example 1. 2x² − 3x + 5 = 0: D = 9 − 40 = −31 < 0, no real roots.

Example 2. 3x² − 4√3 x + 4 = 0: D = 48 − 48 = 0, equal roots, x = 4√3/6 = 2/√3 = 2√3/3.

Example 3. 2x² − 6x + 3 = 0: D = 36 − 24 = 12 > 0, two distinct real roots, x = (6 ± 2√3)/4 = (3 ± √3)/2.

Example 4. x² + 5x + 6 = 0: D = 25 − 24 = 1 > 0, distinct real roots −2 and −3. Note that D being a perfect square means the roots are rational.

Example 5. x² − 6x + 9 = 0: D = 36 − 36 = 0, equal roots x = 3.

Finding a parameter for equal roots. Set D = 0. For 2x² + kx + 3 = 0 to have equal roots: k² − 24 = 0, k = ±2√6. For kx(x − 2) + 6 = 0, that is, kx² − 2kx + 6 = 0: D = 4k² − 24k = 4k(k − 6) = 0, so k = 0 or 6; but k = 0 destroys the quadratic, so k = 6. For x² − 2(k + 1)x + k² = 0 to have equal roots: 4(k + 1)² − 4k² = 0, so (k + 1)² = k², 2k + 1 = 0, k = −1/2. For (k + 1)x² − 2(k − 1)x + 1 = 0 to have equal roots: 4(k − 1)² − 4(k + 1) = 0, so k² − 2k + 1 − k − 1 = 0, k² − 3k = 0, k = 0 or 3.

Finding a parameter for real roots. For x² + kx + 4 = 0 to have real roots: k² − 16 ≥ 0, so k ≥ 4 or k ≤ −4. For 2x² + 3x + k = 0 to have real roots: 9 − 8k ≥ 0, so k ≤ 9/8.

Sum and product of roots. From the formula, the two roots add to −b/a and multiply to c/a, the same relations as for the zeroes of the polynomial. These give quick checks: for 2x² − 7x + 3 = 0 the roots 3 and 1/2 add to 7/2 = −b/a and multiply to 3/2 = c/a.

Reading D in word problems. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? Length x, breadth 40 − x, area x(40 − x) = 400, so x² − 40x + 400 = 0, D = 1600 − 1600 = 0: yes, equal roots x = 20, a square park of side 20 m. Is it possible for two friends to have ages summing to 20 and product 112 four years ago? Ages x and 20 − x, (x − 4)(16 − x) = 112 gives x² − 20x + 112 = 0, D = 400 − 448 < 0: not possible. The discriminant answers possibility questions before any solving.

Number problems translate naturally into quadratics because they involve products or squares. The procedure: name the unknown, write the condition, form the standard-form equation, solve, and check that each root fits the description (positive, integer, and so on).

Consecutive integers. Find two consecutive positive integers whose product is 306. Let them be x and x + 1: x(x + 1) = 306, x² + x − 306 = 0. Split: 18 and −17. (x + 18)(x − 17) = 0, x = 17 or −18. Reject −18 (not positive). The integers are 17 and 18. Check: 17 × 18 = 306.

Consecutive odd numbers. The sum of the squares of two consecutive odd positive integers is 290. Let them be x and x + 2: x² + (x + 2)² = 290, 2x² + 4x + 4 = 290, x² + 2x − 143 = 0, (x + 13)(x − 11) = 0, x = 11. The numbers are 11 and 13. Check: 121 + 169 = 290.

Two numbers with a given sum and sum of squares. The sum of two numbers is 27 and their product is 182. Let them be x and 27 − x: x(27 − x) = 182, x² − 27x + 182 = 0, (x − 13)(x − 14) = 0. The numbers are 13 and 14. Here both roots give the same pair.

Difference of squares. The difference of squares of two numbers is 180 and the square of the smaller is 8 times the larger. Let the larger be x and the smaller y with y² = 8x. Then x² − 8x = 180, x² − 8x − 180 = 0, (x − 18)(x + 10) = 0, x = 18 or −10. If x = 18, y² = 144, y = ±12. If x = −10, y² = −80, impossible. So the numbers are 18 and 12, or 18 and −12.

Reciprocals. The sum of a number and its reciprocal is 10/3. x + 1/x = 10/3, 3x² − 10x + 3 = 0, (3x − 1)(x − 3) = 0, x = 3 or 1/3. Both work (they are each other's reciprocals).

Marbles. John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the marbles they now have is 124. Let John have x; Jivanti has 45 − x. (x − 5)(40 − x) = 124, so −x² + 45x − 200 = 124, x² − 45x + 324 = 0, (x − 9)(x − 36) = 0. John had 9 and Jivanti 36, or John 36 and Jivanti 9.

Toys. A cottage industry produces a certain number of toys in a day. The cost of production of each toy was found to be 55 minus the number of toys produced, and the total cost on a day was Rs 750. x(55 − x) = 750, x² − 55x + 750 = 0, (x − 25)(x − 30) = 0. 25 or 30 toys; both are valid (cost per toy 30 or 25).

Three consecutive integers. The product of two consecutive... rather: the sum of the squares of three consecutive natural numbers is 149. x² + (x + 1)² + (x + 2)² = 149, 3x² + 6x + 5 = 149, x² + 2x − 48 = 0, (x + 8)(x − 6) = 0, x = 6. The numbers are 6, 7, 8. Check: 36 + 49 + 64 = 149.

Always finish by naming the rejected root and the reason. The examiner looks for that line.

Areas, perimeters and Pythagoras produce quadratics because area is a product and Pythagoras involves squares.

Rectangular plot. The area of a rectangular plot is 528 m² and its length is one more than twice its breadth. Breadth x, length 2x + 1: x(2x + 1) = 528, 2x² + x − 528 = 0. a × c = −1056, b = 1: numbers 33 and −32. 2x² + 33x − 32x − 528 = x(2x + 33) − 16(2x + 33) = (2x + 33)(x − 16) = 0. x = 16 (reject −33/2). Breadth 16 m, length 33 m. Check: 16 × 33 = 528.

Rectangular field and diagonal. The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides. Shorter x, longer x + 30, diagonal x + 60. By Pythagoras: x² + (x + 30)² = (x + 60)². 2x² + 60x + 900 = x² + 120x + 3600. x² − 60x − 2700 = 0. (x − 90)(x + 30) = 0. x = 90. Sides 90 m and 120 m, diagonal 150 m. Check: 8100 + 14400 = 22500 = 150².

Right triangle. The altitude of a right triangle is 7 cm less than its base. The hypotenuse is 13 cm. Base x, altitude x − 7: x² + (x − 7)² = 169. 2x² − 14x + 49 = 169. x² − 7x − 60 = 0. (x − 12)(x + 5) = 0. x = 12. Base 12 cm, altitude 5 cm.

Two squares. The sum of the areas of two squares is 468 m². The difference of their perimeters is 24 m. Sides x and y with 4x − 4y = 24, so x = y + 6. (y + 6)² + y² = 468. 2y² + 12y + 36 = 468. y² + 6y − 216 = 0. (y + 18)(y − 12) = 0. y = 12, x = 18. Sides 18 m and 12 m. Check: 324 + 144 = 468.

Park with a given perimeter and area. A rectangular park has perimeter 80 m and area 400 m². Length x, breadth 40 − x: x(40 − x) = 400, x² − 40x + 400 = 0, (x − 20)² = 0, x = 20. A square of side 20 m. If instead the area were 500 m², x² − 40x + 500 = 0 has D = 1600 − 2000 < 0, so no such park exists.

Pole in a park. A pole is to be erected at a point on the boundary of a circular park of diameter 13 m so that the differences of its distances from two diametrically opposite gates A and B is 7 m. Since the angle in a semicircle is a right angle, the distances x and x + 7 satisfy x² + (x + 7)² = 169. 2x² + 14x + 49 = 169. x² + 7x − 60 = 0. (x + 12)(x − 5) = 0. x = 5. The pole is 5 m from one gate and 12 m from the other. Check: 25 + 144 = 169.

Ladder against a wall. A 10 m ladder reaches a window; if the foot is moved 2 m further from the wall the top comes 4 m lower... such problems give two Pythagoras equations that reduce to a linear one, or a single quadratic depending on the data. Setting up carefully is the skill.

Lengths cannot be negative; reject negative roots explicitly. Also verify with Pythagoras or the area at the end, because a sign slip in expansion often gives a plausible but wrong length.

Speed and work problems produce quadratics when a time difference is expressed as a difference of two fractions with the unknown in the denominators.

Train with reduced speed. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less. Find the speed. Let the speed be x km/h. Time at x is 360/x; at x + 5 it is 360/(x + 5). The difference is 1 hour: 360/x − 360/(x + 5) = 1. Multiply by x(x + 5): 360(x + 5) − 360x = x(x + 5), so 1800 = x² + 5x, x² + 5x − 1800 = 0. Split: 45 and −40. (x + 45)(x − 40) = 0. x = 40 (reject −45). Speed 40 km/h. Check: 360/40 = 9 h, 360/45 = 8 h, difference 1 h.

Express train. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. The average speed of the express train is 11 km/h more than that of the passenger train. Passenger speed x: 132/x − 132/(x + 11) = 1. 132(x + 11) − 132x = x(x + 11). 1452 = x² + 11x. x² + 11x − 1452 = 0. Split: 44 and −33. (x + 44)(x − 33) = 0. x = 33. Passenger 33 km/h, express 44 km/h. Check: 4 h and 3 h.

Aeroplane. An aeroplane left 30 minutes later than its scheduled time and, in order to reach its destination 1500 km away in time, had to increase its speed by 250 km/h from its usual speed. Usual speed x: 1500/x − 1500/(x + 250) = 1/2. 1500 × 250 × 2 = x(x + 250). x² + 250x − 750000 = 0. D = 62500 + 3000000 = 3062500 = 1750². x = (−250 + 1750)/2 = 750. Usual speed 750 km/h.

Boat in a stream. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Stream speed x: 24/(18 − x) − 24/(18 + x) = 1. 24(18 + x) − 24(18 − x) = (18 − x)(18 + x). 48x = 324 − x². x² + 48x − 324 = 0. Split: 54 and −6. (x + 54)(x − 6) = 0. x = 6. Stream 6 km/h. Check: 24/12 = 2 h, 24/24 = 1 h.

Two taps. Two water taps together can fill a tank in 9 3/8 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time each takes. Smaller tap x hours, larger x − 10. Per hour: 1/x + 1/(x − 10) = 8/75. Multiply by 75x(x − 10): 75(x − 10) + 75x = 8x(x − 10). 150x − 750 = 8x² − 80x. 8x² − 230x + 750 = 0. Divide by 2: 4x² − 115x + 375 = 0. D = 13225 − 6000 = 7225 = 85². x = (115 ± 85)/8 = 25 or 15/4. Reject 15/4 because then x − 10 is negative. Smaller tap 25 hours, larger 15 hours. Check: 1/25 + 1/15 = 8/75.

Work by two people. A takes 6 days less than B to finish a piece of work; together they take 4 days. B takes x days, A takes x − 6: 1/x + 1/(x − 6) = 1/4. 4(x − 6) + 4x = x(x − 6). 8x − 24 = x² − 6x. x² − 14x + 24 = 0. (x − 12)(x − 2) = 0. x = 12 (x = 2 makes A's time negative). B 12 days, A 6 days.

Clearing denominators is where errors occur; multiply every term by the full product of denominators, and keep the equation balanced. Then reject any root that makes a speed, a time or a difference negative.

Ages. Rohan's mother is 26 years older than him. The product of their ages three years from now will be 360. Rohan's age x: (x + 3)(x + 29) = 360. x² + 32x + 87 = 360. x² + 32x − 273 = 0. Split: 39 and −7. (x + 39)(x − 7) = 0. x = 7. Rohan is 7, his mother 33. Check: 10 × 36 = 360.

The sum of the ages of a father and son is 45 years. Five years ago the product of their ages was 124. Son x, father 45 − x: (x − 5)(40 − x) = 124. −x² + 45x − 200 = 124. x² − 45x + 324 = 0. (x − 9)(x − 36) = 0. x = 9 (36 would make the father 9). Son 9, father 36.

A girl is twice as old as her sister. Four years hence the product of their ages will be 160. Sister x, girl 2x: (x + 4)(2x + 4) = 160. 2x² + 12x + 16 = 160. x² + 6x − 72 = 0. (x + 12)(x − 6) = 0. x = 6. Sister 6, girl 12.

Shopping and pricing. A shopkeeper buys a number of books for Rs 80. If he had bought 4 more books for the same amount, each book would have cost Rs 1 less. Number of books x: 80/x − 80/(x + 4) = 1. 80(x + 4) − 80x = x(x + 4). 320 = x² + 4x. x² + 4x − 320 = 0. (x + 20)(x − 16) = 0. x = 16 books.

Class picnic. Some students planned a picnic. The budget for food was Rs 480. As 8 of them failed to go, the cost of food for each member increased by Rs 10. Original number x: 480/(x − 8) − 480/x = 10. 480x − 480(x − 8) = 10x(x − 8). 3840 = 10x² − 80x. x² − 8x − 384 = 0. (x − 24)(x + 16) = 0. x = 24. Those who went: 16, paying Rs 30 each.

Articles. The cost of producing x articles is 2x² + 3x + 5 rupees... if the selling price per article is (3x + 2)... such profit problems set revenue minus cost equal to a stated profit and give a quadratic.

Bird flock (a classical problem). One-fourth of a flock of pigeons... setting the sum of fractional parts and a remainder equal to the whole gives a quadratic in the square root of the flock size; substitute y = √x to obtain y² in x and solve. These appear as challenge items.

Multiple choice patterns. Which of the following is a root of 2x² − 5x + 3 = 0? Test each option by substitution: x = 1 gives 2 − 5 + 3 = 0, so 1 is a root; the other is 3/2 by product of roots 3/2.

Choosing the unknown wisely. Let x be the quantity the question asks for, unless another choice gives a simpler equation. In the picnic problem, choosing the original number gives a cleaner equation than choosing the number who went. In age problems, choosing the younger person's age keeps the numbers small.

Checking admissibility. Ages, counts and prices must be positive, and counts must be whole numbers; a root such as 15/4 books signals a rejected root. Also check that derived quantities (a father's age, a reduced group) are sensible.

Some equations are not quadratic at first sight but become quadratic after simplification or substitution.

Fractions with x in denominators. 1/(x − 3) + 1/(x + 3) = 2/... let us use: x/(x + 1) + (x + 1)/x = 34/15, x ≠ 0, −1. Put y = x/(x + 1); then (x + 1)/x = 1/y, and y + 1/y = 34/15, so 15y² − 34y + 15 = 0, (5y − 3)(3y − 5) = 0, y = 3/5 or 5/3. If x/(x + 1) = 3/5: 5x = 3x + 3, x = 3/2. If x/(x + 1) = 5/3: 3x = 5x + 5, x = −5/2. Both admissible.

Sum of reciprocals. 1/(x + 4) − 1/(x − 7) = 11/30: shown earlier to give x² − 3x + 2 = 0, x = 1, 2.

Another fraction equation. (x − 1)/(x − 2) + (x − 3)/(x − 4) = 10/3, x ≠ 2, 4. Combine: [(x − 1)(x − 4) + (x − 3)(x − 2)]/[(x − 2)(x − 4)] = 10/3. Numerator: x² − 5x + 4 + x² − 5x + 6 = 2x² − 10x + 10. Denominator: x² − 6x + 8. Cross multiply: 6x² − 30x + 30 = 10x² − 60x + 80, so 4x² − 30x + 50 = 0, 2x² − 15x + 25 = 0, (2x − 5)(x − 5) = 0, x = 5/2 or 5.

Quadratic in x². x⁴ − 5x² + 4 = 0. Put y = x²: y² − 5y + 4 = 0, (y − 1)(y − 4) = 0, y = 1 or 4. So x² = 1 gives x = ±1, x² = 4 gives x = ±2. Four real roots.

Square root equations. √(2x + 9) + x = 13. Isolate the root: √(2x + 9) = 13 − x. Square: 2x + 9 = 169 − 26x + x². x² − 28x + 160 = 0. (x − 8)(x − 20) = 0. Check x = 8: √25 + 8 = 13, valid. Check x = 20: √49 + 20 = 27 ≠ 13, reject. Squaring can introduce false roots, so checking is compulsory.

Equations with x + 1/x. x² + 1/x² = 7... use (x + 1/x)² = x² + 1/x² + 2 = 9, so x + 1/x = ±3, then x² ∓ 3x + 1 = 0, giving x = (3 ± √5)/2 or (−3 ± √5)/2.

Product form. (x + 1)(x + 2)(x + 3)(x + 4) = 24... pair as [(x + 1)(x + 4)][(x + 2)(x + 3)] = (x² + 5x + 4)(x² + 5x + 6) = 24. Put y = x² + 5x + 5: (y − 1)(y + 1) = 24, y² = 25, y = ±5. y = 5: x² + 5x = 0, x = 0 or −5. y = −5: x² + 5x + 10 = 0, D < 0, no real root. Real roots 0 and −5. Check: 1 × 2 × 3 × 4 = 24 and (−4)(−3)(−2)(−1) = 24.

Principle. Look for a repeated block, name it, solve the quadratic in the block, then return to x. State the excluded values from any denominators and reject roots that were introduced by squaring. These reduce to routine work once the block is spotted, and they are the most likely source of a four-mark challenge question.

The roots of ax² + bx + c = 0 are the x-coordinates of the points where the parabola y = ax² + bx + c meets the x-axis. This picture connects the three methods and the discriminant.

Vertex form. Completing the square gives y = a(x + b/2a)² + (c − b²/4a) = a(x + b/2a)² − D/4a. The vertex is at x = −b/2a with y = −D/4a. For a > 0, the vertex is the lowest point; the parabola reaches the x-axis exactly when −D/4a ≤ 0, that is, D ≥ 0. So D > 0 gives two crossings, D = 0 gives a touch at the vertex, D < 0 gives no contact.

Example. y = x² − 4x + 3 = (x − 2)² − 1. Vertex (2, −1), below the axis; D = 16 − 12 = 4 > 0; crossings at x = 2 ± 1 = 1 and 3. y = x² − 4x + 4 = (x − 2)²: vertex (2, 0), on the axis; D = 0; the root 2 is repeated. y = x² − 4x + 5 = (x − 2)² + 1: vertex (2, 1), above the axis; D = −4; no real root.

Symmetry of roots. The two roots are symmetric about the vertex: α = −b/2a − √D/2a and β = −b/2a + √D/2a. Their midpoint is −b/2a, the axis of symmetry, and their distance apart is √D/|a|. So the sum is −b/a and the product, by expanding (x − α)(x − β), is c/a.

Drawing to solve. To solve x² − x − 6 = 0 graphically, plot y = x² − x − 6 from a table: x = −3, y = 6; −2, 0; −1, −4; 0, −6; 1, −6; 2, −4; 3, 0; 4, 6. The curve crosses at x = −2 and x = 3, the roots. Algebra confirms: (x − 3)(x + 2) = 0.

Solving a system with a parabola and a line. Where does y = x² meet y = x + 2? Set x² = x + 2, x² − x − 2 = 0, (x − 2)(x + 1) = 0, x = 2 or −1; points (2, 4) and (−1, 1). So a quadratic equation also arises from intersecting a parabola with a line, and D tells whether the line cuts, touches or misses the parabola. The line y = x − 1 with y = x²: x² − x + 1 = 0, D = −3, so the line misses the parabola.

Maximum and minimum from the vertex. Because the vertex is the extreme point, a quadratic expression has a least value (a > 0) or greatest value (a < 0) of −D/4a at x = −b/2a. The expression x² − 6x + 13 = (x − 3)² + 4 has minimum 4 at x = 3. A farmer with 40 m of fencing for a rectangle against a wall maximises area x(40 − 2x) = −2x² + 40x at x = 10, area 200 m². This is the beginning of optimisation, met properly in Intermediate mathematics.

In the examination the graphical view appears as reasoning questions: given D, sketch the parabola; given a sketch, state the sign of D; find the vertex and hence the number of roots. Keep the vertex formula and the three pictures in mind.

This chapter defined the quadratic equation ax² + bx + c = 0, gave three methods of solution, introduced the discriminant, and applied all of it to word problems.

Facts to memorise. Standard form and the condition a ≠ 0. Zero product rule. The completing-the-square step: add (b/2a)² after dividing by a. The quadratic formula x = [−b ± √(b² − 4ac)]/2a. D = b² − 4ac and its three cases. Sum of roots −b/a, product c/a. Vertex at −b/2a.

One-mark questions. Is a given equation quadratic? Write the discriminant of 2x² − 3x + 1 = 0. State the nature of roots of x² + 4 = 0. Find k so that x² + kx + 9 = 0 has equal roots. Check whether 2 is a root of x² − 5x + 6 = 0. Form the quadratic equation for a plot whose length is twice its breadth and area 800.

Two-mark questions. Solve by factorisation. Find the roots using the formula. Find the nature of the roots. Find the value of a parameter for equal roots. Solve an equation with simple fractions.

Four-mark questions. Solve by completing the square, showing each step. A word problem on speed, area, ages or numbers: form, solve, reject, answer, check. Determine whether a described situation is possible using D and, if so, find the dimensions. Solve a reducible equation.

Recurring errors. Forgetting the ± when taking square roots. Dividing only part of the numerator by 2a. Writing D with the wrong sign of c. Accepting a negative length or age. Not checking roots after squaring. Giving only one root when two are admissible (as with marbles 9 and 36). Forgetting excluded values in fraction equations.

Answer layout for word problems. Let x be ... (with unit). Then ... (express the other quantities). Equation: ... . Standard form: ... . Solve: ... . Roots: ... . Reject ... because ... . Answer: ... . Check: ... .

Revision routine. Daily: factorise three, complete the square on one, apply the formula to two including one with irrational roots, compute D for three and state the nature, solve one word problem from a different family each day. Weekly: one full mock of the chapter's likely questions in 40 minutes.

Looking ahead. Progressions uses quadratics to find the number of terms when a sum is given. Coordinate geometry uses them when a distance is known. Trigonometry and mensuration produce them in height and volume problems. In the Intermediate course, the discriminant extends to complex roots and the theory of equations, and quadratic functions are the first examples in calculus of maxima and minima.