Chapter 9 — Quadratic Equations Optional

The routine exercises of the chapter give you a quadratic equation, or a situation so simple that the equation almost writes itself. The optional exercise is different in one way only: the quadratic is hidden. The situation is described in words, sometimes in two or three sentences, and the equation appears only after you have chosen an unknown, expressed every quantity in the problem in terms of that unknown, and written the one relation that the problem guarantees. Once the equation is on paper, the solving is the same factorisation, completing the square or formula that you already know.

So the skill being tested is translation, and a good translation has three parts. First, name the unknown precisely: not "the number" but "the tens digit", not "the speed" but "the speed of the first train in km/h". Second, write every other quantity in terms of it before touching any equation: if the tens digit is x and the product of the digits is 8, the units digit is 8/x and the number is 10x + 8/x. Third, find the sentence that gives an equation. Every word problem contains exactly one sentence that equates two quantities — "the cost remains the same", "they are 50 km apart", "the digits are reversed" — and that sentence is where the equation comes from.

After solving, there is a fourth step that is often forgotten: check the roots against the situation. A quadratic gives two roots, and a word problem usually accepts only one. A number of points cannot be −4; a digit cannot be 16; a speed cannot be negative; a length cannot exceed the whole. Writing "x = 5 or x = −4; since x is a number of points, x = 5" is part of the answer and earns marks.

Finally, the optional exercise also has purely algebraic problems: quadratics with irrational roots, equations that become quadratic after a substitution such as y = x + 1/x, and questions where a coefficient k must be found so that the roots are equal. These test whether you understand the structure of the quadratic, not just the mechanics. This chapter takes each family of problem in turn, works one or two examples fully, and points out the trap in each.

The first problem of the optional exercise is a counting problem. Some points are marked on a plane so that no three of them lie on one straight line. Every point is joined to every other point by a line segment, and the total number of segments is 10. How many points were marked?

Let the number of points be n. From any one point, segments go to each of the other n − 1 points, so each point has n − 1 segments leaving it, giving n(n − 1) segment-ends. But every segment has two ends and has been counted twice, once from each of its endpoints. So the number of distinct segments is n(n − 1)/2. This is the same formula that counts handshakes in a room of n people, or matches in a league where every team plays every other team once.

The equation is n(n − 1)/2 = 10, so n(n − 1) = 20, so n² − n − 20 = 0. To factorise, look for two numbers with product −20 and sum −1: they are −5 and 4. So (n − 5)(n + 4) = 0, and n = 5 or n = −4. A number of points cannot be negative, so n = 5. Check: five points give 5 × 4/2 = 10 segments. Correct.

The condition "no three collinear" matters. If three points were on one line, the segment joining the outer two would pass through the middle one and the segments would overlap, so the count of distinct segments would be less than n(n − 1)/2. The problem includes the condition precisely so that the formula holds.

The same formula answers a family of questions. If 45 handshakes were exchanged when everyone shook hands with everyone else once, n(n − 1)/2 = 45 gives n² − n − 90 = 0, (n − 10)(n + 9) = 0, so 10 people. If a league of teams played 66 matches with each pair meeting once, n² − n − 132 = 0, (n − 12)(n + 11) = 0, so 12 teams. If the number of diagonals of a polygon is asked, the formula is n(n − 3)/2 because the n sides are not diagonals: a polygon with 20 diagonals satisfies n² − 3n − 40 = 0, (n − 8)(n + 5) = 0, so it is an octagon.

In the examination, write the derivation of n(n − 1)/2 in one line, the equation, the factorisation, both roots, and the rejection of the negative root. Each of these steps carries marks.

A two-digit number is such that the product of its digits is 8. When 18 is added to the number, the digits are reversed. Find the number.

The key is to represent a two-digit number correctly. If the tens digit is x and the units digit is y, the number is 10x + y, not xy. The reversed number is 10y + x. Here the product of the digits is 8, so y = 8/x, and the number is 10x + 8/x.

The condition "adding 18 reverses the digits" gives the equation: (10x + 8/x) + 18 = 10(8/x) + x. Multiply every term by x to clear the fraction: 10x² + 8 + 18x = 80 + x². Bring everything to one side: 9x² + 18x − 72 = 0. Divide by 9: x² + 2x − 8 = 0. Factorise: two numbers with product −8 and sum 2 are 4 and −2, so (x + 4)(x − 2) = 0, giving x = 2 or x = −4. A digit cannot be negative, so x = 2, y = 8/2 = 4, and the number is 24. Check: 24 + 18 = 42, which is 24 reversed, and 2 × 4 = 8. Correct.

Notice that the working produced a term 8/x, and clearing it multiplied the whole equation by x. This is legitimate because x is a digit and cannot be 0 (the tens digit of a two-digit number is never 0). Always state this when you multiply through by an unknown.

Variants use the sum of digits, a difference instead of a sum, or a subtraction reversing the digits. If the digits sum to 9 and the number exceeds its reverse by 27, then y = 9 − x and (10x + y) − (10y + x) = 27 gives 9(x − y) = 27, x − y = 3, and this is linear: x = 6, y = 3, number 63. The problem only becomes quadratic when the product of the digits is given, because y = P/x introduces the reciprocal. If the product is 12 and adding 36 reverses the number, then 10x + 12/x + 36 = 120/x + x, so 10x² + 12 + 36x = 120 + x², 9x² + 36x − 108 = 0, x² + 4x − 12 = 0, (x + 6)(x − 2) = 0, x = 2, and the number is 26; check 26 + 36 = 62.

A useful shortcut: when adding k reverses the digits, 10y + x − (10x + y) = k gives 9(y − x) = k, so k must be a multiple of 9 and y − x = k/9. For k = 18, y − x = 2; with xy = 8 the pair (2, 4) is found at once. Use this to check the answer, but show the quadratic in the examination.

A piece of cloth costs ₹200. If the piece were 5 m longer and each metre cost ₹2 less, the cost of the piece would remain unchanged. How long is the piece and what is the rate per metre?

Let the length be x metres. Then the rate is 200/x rupees per metre, because total cost = length × rate. In the changed situation the length is x + 5 and the rate is 200/x − 2, and the total is still 200. So the equation is (x + 5)(200/x − 2) = 200.

Expand: 200 − 2x + 1000/x − 10 = 200. Simplify: −2x + 1000/x − 10 = 0. Multiply by x (x ≠ 0 since the length is positive): −2x² + 1000 − 10x = 0, that is, 2x² + 10x − 1000 = 0, or x² + 5x − 500 = 0. Factorise: two numbers with product −500 and sum 5 are 25 and −20. So (x + 25)(x − 20) = 0, x = 20 or x = −25. Length is positive, so x = 20 m and the rate is 200/20 = ₹10 per metre. Check: 25 m at ₹8 per metre costs ₹200. Correct.

This structure — a fixed total, one factor rising by a constant while the other falls by a constant — appears in many disguises: a shopkeeper buying books for ₹80 who could buy 4 more if each cost ₹1 less; a group sharing ₹1200 equally where 3 more members would reduce each share by ₹20; a journey of fixed distance where a higher speed by 10 km/h saves 1 hour. The equation always has the form (x + p)(T/x − q) = T, which reduces to qx² + pqx − pT = 0. The middle term is easy to get wrong; expand carefully and collect like terms before multiplying through.

The books example: let the price be x, so 80/x books are bought. Then 80/(x − 1) = 80/x + 4. Multiply by x(x − 1): 80x = 80(x − 1) + 4x(x − 1), so 80x = 80x − 80 + 4x² − 4x, so 4x² − 4x − 80 = 0, x² − x − 20 = 0, (x − 5)(x + 4) = 0, price ₹5 and 16 books. Whichever quantity you choose as the unknown, the answer is the same; choose the one that gives simpler algebra, usually the one that appears in the "per unit" phrase.

Two trains leave a railway station at the same time. The first travels due west and the second due north. The second train travels 5 km/h faster than the first. After two hours they are 50 km apart. Find the speed of each train.

Let the speed of the first train be x km/h; then the second train's speed is x + 5 km/h. In two hours the first train covers 2x km westward and the second covers 2(x + 5) = 2x + 10 km northward. West and north are perpendicular, so the station and the two trains form a right-angled triangle with the right angle at the station. The distance between the trains is the hypotenuse.

By Pythagoras: (2x)² + (2x + 10)² = 50². So 4x² + 4x² + 40x + 100 = 2500. Collect: 8x² + 40x − 2400 = 0. Divide by 8: x² + 5x − 300 = 0. Factorise: product −300, sum 5 — the numbers are 20 and −15. So (x + 20)(x − 15) = 0, x = 15 or x = −20. Speed is positive, so the first train travels at 15 km/h and the second at 20 km/h. Check: in two hours they cover 30 km and 40 km, and √(30² + 40²) = √2500 = 50 km. The 3-4-5 triangle scaled by 10.

A cleaner way to write this is to note that the distances in two hours, 2x and 2x + 10, could be renamed: let d = 2x, then d² + (d + 10)² = 2500 gives 2d² + 20d − 2400 = 0, d² + 10d − 1200 = 0, (d + 40)(d − 30) = 0, d = 30 km, so x = 15 km/h. Either route is acceptable.

The same geometry underlies problems about a pole and its shadow, a ladder sliding down a wall, or a boat crossing a river while the current carries it downstream. In each, two perpendicular displacements and the straight-line distance form a right triangle, and if one displacement is expressed in terms of the other, the Pythagoras relation is a quadratic. A pole of height h with a shadow h + 7 and a distance from the top of the pole to the tip of the shadow of 13 m: h² + (h + 7)² = 169, 2h² + 14h − 120 = 0, h² + 7h − 60 = 0, (h + 12)(h − 5) = 0, h = 5 m.

Draw the diagram in the examination. A labelled right triangle showing the station, the westward distance 2x, the northward distance 2x + 10 and the hypotenuse 50 makes the equation self-evident and earns the presentation mark.

Not every quadratic factorises over the integers. When you cannot find two numbers with the right product and sum, the roots are irrational (or not real), and you must use completing the square or the formula and leave the answer in exact surd form. The optional exercise includes equations like x² − 4x − 8 = 0 and 2x² + x − 4 = 0 precisely to test this.

Completing the square rests on the identity (x + p)² = x² + 2px + p². To solve x² − 4x − 8 = 0: move the constant, x² − 4x = 8. Half the coefficient of x is −2, and its square is 4; add 4 to both sides: x² − 4x + 4 = 12, so (x − 2)² = 12. Take square roots: x − 2 = ±√12 = ±2√3, so x = 2 ± 2√3. Both roots are real and irrational; approximately 5.46 and −1.46.

When the leading coefficient is not 1, divide through first. For 2x² + x − 4 = 0: divide by 2, x² + x/2 − 2 = 0, so x² + x/2 = 2. Half of 1/2 is 1/4, square is 1/16. Add: x² + x/2 + 1/16 = 2 + 1/16 = 33/16, so (x + 1/4)² = 33/16, x + 1/4 = ±√33/4, x = (−1 ± √33)/4. The formula gives the same: a = 2, b = 1, c = −4, discriminant 1 + 32 = 33, x = (−1 ± √33)/4.

The quadratic formula x = [−b ± √(b² − 4ac)]/2a is completing the square done once for all. Use it when the numbers are awkward; use completing the square when the question asks for that method, or when the equation is of the form x² + 2px + q = 0 with a small even coefficient. Always simplify the surd: √12 = 2√3, √48 = 4√3, √50 = 5√2, and cancel common factors between the surd term, the rational term and the denominator. For 3x² − 6x + 1 = 0: discriminant 36 − 12 = 24, x = (6 ± √24)/6 = (6 ± 2√6)/6 = (3 ± √6)/3 = 1 ± √6/3.

A frequent error is to add the square only to one side, or to forget to divide by the leading coefficient before halving. Another is to write ±√12 as ±12 or as ±√3. Write each step on its own line and check the final roots by substitution into the original equation, at least approximately: 2 + 2√3 ≈ 5.464, and 5.464² − 4(5.464) − 8 ≈ 29.86 − 21.86 − 8 = 0. Good.

Some equations are not quadratic as written but become quadratic when a suitable expression is renamed as a single variable. Recognising the substitution is the whole art; after it, the solving is routine, and the only extra work is to go back and find the original unknown from each root.

Biquadratic equations such as x⁴ − 13x² + 36 = 0 contain only even powers. Put y = x²: y² − 13y + 36 = 0, (y − 4)(y − 9) = 0, y = 4 or 9. Then x² = 4 gives x = ±2 and x² = 9 gives x = ±3. Four roots: −3, −2, 2, 3. If a value of y is negative, x² = y has no real root and is discarded.

Reciprocal expressions such as (x + 1/x)² − 5(x + 1/x) + 6 = 0 call for y = x + 1/x: y² − 5y + 6 = 0, y = 2 or 3. Then x + 1/x = 2 gives x² − 2x + 1 = 0, x = 1 (repeated), and x + 1/x = 3 gives x² − 3x + 1 = 0, x = (3 ± √5)/2. Note that x + 1/x = 2 has only one root because (x − 1)² = 0.

Square roots: √(2x + 9) + x = 13. Isolate the root, √(2x + 9) = 13 − x, square both sides, 2x + 9 = 169 − 26x + x², so x² − 28x + 160 = 0, (x − 8)(x − 20) = 0, x = 8 or 20. Squaring can introduce false roots, so check: x = 8 gives √25 + 8 = 13, correct; x = 20 gives √49 + 20 = 27 ≠ 13, rejected. Always check after squaring.

Fractions with the unknown in the denominator: 1/(x − 1) + 2/(x − 2) = 6/x. Multiply by x(x − 1)(x − 2): x(x − 2) + 2x(x − 1) = 6(x − 1)(x − 2), so x² − 2x + 2x² − 2x = 6x² − 18x + 12, so 3x² − 4x = 6x² − 18x + 12, so 3x² − 14x + 12 = 0. The discriminant is 196 − 144 = 52, so x = (14 ± 2√13)/6 = (7 ± √13)/3. Neither root is 0, 1 or 2, so both are valid. Values that make a denominator zero must always be excluded.

Exponential forms like 4ˣ − 6·2ˣ + 8 = 0 use y = 2ˣ (since 4ˣ = (2ˣ)²): y² − 6y + 8 = 0, y = 2 or 4, so 2ˣ = 2 gives x = 1 and 2ˣ = 4 gives x = 2.

In every case the procedure is the same: identify the repeated block, name it y, solve the quadratic in y, translate each y back to x, and discard any x that fails the original equation or makes it undefined.

For ax² + bx + c = 0, the quantity D = b² − 4ac is the discriminant, and it decides the nature of the roots without solving: D > 0 gives two distinct real roots, D = 0 gives two equal real roots (one repeated root, −b/2a), and D < 0 gives no real roots. The optional exercise reverses the usual question: a coefficient contains an unknown k, and you must find k so that the roots are of a stated kind.

Equal roots. Find k so that kx² + 2x + 1 = 0 has equal roots. Here a = k, b = 2, c = 1, so D = 4 − 4k. Equal roots need D = 0, so k = 1. Check: x² + 2x + 1 = (x + 1)² = 0, root −1 repeated. For 2x² + kx + 3 = 0: D = k² − 24 = 0, so k = ±2√6. For kx(x − 2) + 6 = 0, first write it as kx² − 2kx + 6 = 0: D = 4k² − 24k = 4k(k − 6) = 0, so k = 0 or k = 6; but k = 0 makes the equation 6 = 0, not quadratic, so k = 6. Always reject the value of k that kills the x² term.

Real and distinct roots. For x² − 2kx + 4 = 0, D = 4k² − 16 > 0 gives k² > 4, so k > 2 or k < −2. No real roots. For the same equation, D < 0 gives −2 < k < 2.

Two unknowns tied by equal roots. If (k + 1)x² − 2(k − 1)x + 1 = 0 has equal roots: D = 4(k − 1)² − 4(k + 1) = 0, so (k − 1)² = k + 1, k² − 2k + 1 = k + 1, k² − 3k = 0, k(k − 3) = 0, k = 0 or 3. Both keep k + 1 ≠ 0, so both are valid. The equal root is then −b/2a = (k − 1)/(k + 1): for k = 0 it is −1, for k = 3 it is 1/2.

Rational versus irrational roots. If a, b, c are rational and D is a perfect square of a rational number, the roots are rational; if D is positive but not a perfect square, the roots are irrational and occur as a conjugate pair p ± √q. Since 2x² + x − 4 = 0 has D = 33, its roots (−1 ± √33)/4 are irrational conjugates. Since x² − 5x + 6 = 0 has D = 1, its roots 2 and 3 are rational.

A problem may state the nature in words: "the equation has a double root", "the graph touches the x-axis", "the parabola does not meet the x-axis". Translate: double root or touching means D = 0; not meeting means D < 0; crossing at two points means D > 0.

Speed problems are the commonest source of quadratics in the SSC paper, because time = distance ÷ speed puts the unknown in a denominator, and a difference of two times gives a difference of two fractions, which clears to a quadratic.

A stream. A motorboat whose speed in still water is 18 km/h takes 1 hour more to go 24 km upstream than to return downstream. Find the speed of the stream. Let the stream's speed be x km/h. Upstream speed is 18 − x, downstream 18 + x. Time upstream is 24/(18 − x), time downstream 24/(18 + x), and the difference is 1: 24/(18 − x) − 24/(18 + x) = 1. Combine: 24[(18 + x) − (18 − x)]/[(18 − x)(18 + x)] = 1, so 24 · 2x = 324 − x², so x² + 48x − 324 = 0. Product −324, sum 48: the numbers are 54 and −6. (x + 54)(x − 6) = 0, x = 6 km/h. Check: upstream 24/12 = 2 h, downstream 24/24 = 1 h, difference 1 h. Correct.

A delayed train. 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: 360/x − 360/(x + 5) = 1, so 360 · 5 = x(x + 5), x² + 5x − 1800 = 0, (x + 45)(x − 40) = 0, x = 40 km/h. Check: 9 h versus 8 h.

A journey in two parts. A person travels 300 km, partly by train at 60 km/h and partly by bus at 40 km/h; the general pattern for the optional exercise is when the speeds themselves are unknown and related, for instance "the bus is 20 km/h slower than the train and the total time was 6 hours with 120 km by bus": 180/x + 120/(x − 20) = 6, multiply by x(x − 20): 180(x − 20) + 120x = 6x(x − 20), 300x − 3600 = 6x² − 120x, 6x² − 420x + 3600 = 0, x² − 70x + 600 = 0, (x − 60)(x − 10) = 0. Here x = 10 would make the bus speed −10, so x = 60 km/h.

The general shape is d/x − d/(x + p) = t, which simplifies to x² + px − dp/t = 0. Memorising this is not needed; what is needed is the habit of writing the two times as fractions, subtracting the smaller from the larger, and multiplying through by the product of the denominators. The negative root is always rejected, and if the root makes any speed in the problem negative or zero, it too is rejected.

Work problems become quadratic when two workers' times are related and their combined time is given. The principle is that rates add: if one person does a job in a days, they do 1/a of it per day; if two people work together, their daily fractions add and the whole job is done in the time for which the sum equals 1.

Two taps. Two water taps together can fill a tank in 9⅜ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time each tap would take alone. Let the smaller tap take x hours; the larger takes x − 10. Together they fill 1/x + 1/(x − 10) of the tank per hour, and this equals 1/(75/8) = 8/75. So 1/x + 1/(x − 10) = 8/75. Combine: (2x − 10)/[x(x − 10)] = 8/75. Cross-multiply: 75(2x − 10) = 8x(x − 10), 150x − 750 = 8x² − 80x, 8x² − 230x + 750 = 0, divide by 2: 4x² − 115x + 375 = 0. Discriminant: 115² − 4·4·375 = 13225 − 6000 = 7225 = 85². x = (115 ± 85)/8 = 25 or 3.75. If x = 3.75, the larger tap takes −6.25 hours, impossible. So the smaller tap takes 25 hours and the larger 15 hours. Check: 1/25 + 1/15 = (3 + 5)/75 = 8/75. Correct.

Two workers. A takes 6 days less than B to finish a piece of work; together they finish it in 4 days. Let B take x days; A takes x − 6. Then 1/x + 1/(x − 6) = 1/4, so 4(2x − 6) = x(x − 6), 8x − 24 = x² − 6x, x² − 14x + 24 = 0, (x − 12)(x − 2) = 0. x = 2 gives A −4 days, rejected; so B takes 12 days and A 6 days.

The rejection of one root is not a formality here: the rejected root is often positive, and only the constraint that the other worker's time must be positive rules it out. Always compute the second quantity from each root before deciding.

A pipe that empties a tank has a negative rate. If a filling pipe takes x hours and an emptying pipe takes x + 5 hours, and with both open the tank fills in 30 hours: 1/x − 1/(x + 5) = 1/30, so 30 · 5 = x(x + 5), x² + 5x − 150 = 0, (x + 15)(x − 10) = 0, x = 10 hours.

Because area involves a product of two lengths, any problem in which the lengths are related and the area is given is quadratic. The optional exercise adds a twist: the relation is sometimes through the perimeter or the diagonal rather than a simple difference.

Rectangle with given perimeter and area. A rectangular field has perimeter 82 m and area 400 m². Find its sides. Let the length be x; then the breadth is 41 − x (half the perimeter minus the length). Area: x(41 − x) = 400, so x² − 41x + 400 = 0, (x − 25)(x − 16) = 0. The sides are 25 m and 16 m; both roots are valid because they simply exchange length and breadth.

Right triangle with given hypotenuse. The hypotenuse of a right triangle is 25 cm and the difference of the other two sides is 5 cm. Let the shorter side be x; the other is x + 5. x² + (x + 5)² = 625, 2x² + 10x − 600 = 0, x² + 5x − 300 = 0, (x + 20)(x − 15) = 0, x = 15. Sides 15, 20, 25 cm.

Path around a garden. A rectangular lawn 20 m by 15 m has a path of uniform width x around it, and the area of the path is 246 m². The outer rectangle is (20 + 2x) by (15 + 2x). Path area = outer − inner: (20 + 2x)(15 + 2x) − 300 = 246, so 300 + 40x + 30x + 4x² − 300 = 246, 4x² + 70x − 246 = 0, 2x² + 35x − 123 = 0. Discriminant 1225 + 984 = 2209 = 47². x = (−35 ± 47)/4 = 3 or −20.5. Width 3 m.

Altitude of a triangle. The altitude of a right triangle is 7 cm less than its base, and the hypotenuse is 13 cm. Base x, altitude x − 7: x² + (x − 7)² = 169, 2x² − 14x − 120 = 0, x² − 7x − 60 = 0, (x − 12)(x + 5) = 0, base 12 cm, altitude 5 cm.

Sum of areas of two squares. The sum of the areas of two squares is 468 m², and the difference of their perimeters is 24 m. If the sides are x and y, then 4x − 4y = 24 gives x = y + 6, and (y + 6)² + y² = 468, 2y² + 12y − 432 = 0, y² + 6y − 216 = 0, (y + 18)(y − 12) = 0, y = 12 m, x = 18 m.

In all these, sketch the figure, mark the unknown, and write the area or Pythagoras relation directly from the sketch. The common slip is in the path problem — forgetting that the width is added on both sides, so 2x, not x.

Age problems are linear unless a product or a square of ages is mentioned; number problems are quadratic whenever a reciprocal or a square appears. The optional exercise chooses exactly those versions.

Product of ages. The sum of the ages of a father and his son is 45 years. Five years ago the product of their ages was 124. Let the son's present age be x; the father's is 45 − x. Five years ago they were x − 5 and 40 − x, and (x − 5)(40 − x) = 124. Expand: 40x − x² − 200 + 5x = 124, so −x² + 45x − 324 = 0, x² − 45x + 324 = 0. Product 324, sum 45: 36 and 9. (x − 36)(x − 9) = 0. If the son were 36, the father would be 9, absurd; so the son is 9 and the father 36. Check: four years ago they were 4 and 31, and 4 × 31 = 124. Correct. Here both roots are positive and the rejection uses the relation between the two people.

A reciprocal. The sum of a number and its reciprocal is 10/3. Let the number be x: x + 1/x = 10/3, multiply by 3x: 3x² + 3 = 10x, 3x² − 10x + 3 = 0, (3x − 1)(x − 3) = 0, x = 3 or 1/3. Both are valid: 3 + 1/3 = 10/3 and 1/3 + 3 = 10/3. Two answers, and both must be given.

Sum of squares of consecutive numbers. The sum of the squares of two consecutive odd numbers is 394. Let them be x and x + 2: x² + (x + 2)² = 394, 2x² + 4x − 390 = 0, x² + 2x − 195 = 0, (x + 15)(x − 13) = 0. x = 13 gives 13 and 15; x = −15 gives −15 and −13. If the problem says "positive", give 13 and 15; if not, both pairs are correct.

Rohan's mother. Rohan's mother is 26 years older than him. The product of their ages three years from now will be 360. Let Rohan be x; mother x + 26; in three years (x + 3)(x + 29) = 360, x² + 32x + 87 = 360, x² + 32x − 273 = 0, (x + 39)(x − 7) = 0, x = 7 years.

A fraction. The denominator of a fraction is one more than twice its numerator. If the sum of the fraction and its reciprocal is 2 16/21, find the fraction. Numerator x, denominator 2x + 1: x/(2x + 1) + (2x + 1)/x = 58/21. Combine: [x² + (2x + 1)²]/[x(2x + 1)] = 58/21, so 21(5x² + 4x + 1) = 58(2x² + x), 105x² + 84x + 21 = 116x² + 58x, 11x² − 26x − 21 = 0, (11x + 7)(x − 3) = 0, x = 3, fraction 3/7. Check: 3/7 + 7/3 = (9 + 49)/21 = 58/21. Correct.

Facing an unseen problem, the question is never "which formula" but "what kind of relation is hidden here". A short decision routine covers the whole optional exercise.

Step 1 — find the product or the square. Ask where two unknown quantities multiply. Area is length × breadth. Total cost is quantity × rate. Distance is speed × time. A two-digit number involves the product of its digits only when that product is given. Segments among points involve n(n − 1). If nothing multiplies and nothing is squared, the problem is linear, and forcing a quadratic on it is an error.

Step 2 — pick the unknown that appears in a "per" phrase. Speed per hour, cost per metre, days per job. Choosing that quantity as x usually puts the other quantities as simple fractions, and the difference of two fractions with a common numerator is the cleanest route to a quadratic: d/x − d/(x + p) = t.

Step 3 — write the equation, then choose the solving method. If the constant term factorises nicely and the coefficients are small integers, try factorisation by splitting the middle term; look for two numbers whose product is ac and whose sum is b. If the discriminant is a perfect square but the numbers are large (as with 4x² − 115x + 375 = 0, where D = 85²), the formula is faster than hunting for factors. If the discriminant is not a perfect square, the roots are irrational: use the formula or complete the square and leave surds exact. If the question says "by completing the square", do that and show the added square on both sides.

Step 4 — check and reject. Substitute each root, at least mentally, into the original relation. Reject negative lengths, speeds, ages and counts; reject digits outside 0–9; reject roots that make a denominator zero or another quantity negative; after squaring, reject roots that fail the unsquared equation. State the rejection in words.

Presentation for marks. A four-mark answer has: the unknown declared with its unit; the other quantities in terms of it; the equation with a one-line reason ("since the cost is unchanged"); the simplified standard form; the roots; the rejection; the final answer with units; and a check. A diagram for geometry and motion problems. Never skip the standard form — the examiner looks for ax² + bx + c = 0 before the roots.

Finally, remember what the optional exercise is for. It is not examined for the pass mark; it exists so that the student who wants to score high, or to go on to Intermediate mathematics, has met the real shape of quadratic problems: a relation hidden in words, an equation to be built, two roots to be judged. That is the whole of the chapter, and the whole of algebra as it is actually used.