Chapter 4 — Quadratic Equations

In the last chapter we studied the quadratic polynomial ax² + bx + c. When such a polynomial is set equal to zero we get a quadratic equation: ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0. This is the standard form. The condition a ≠ 0 is essential: if a were zero the x² term would vanish and the equation would be linear. The coefficients b and c may be zero: x² − 9 = 0 (b = 0) and 3x² + 5x = 0 (c = 0) are quadratic equations, and so is x² = 0.

The word quadratic comes from the Latin quadratus, meaning square, because the highest power is a square. Quadratic equations were solved in Babylon four thousand years ago, by Indian mathematicians such as Brahmagupta (598 CE) and Sridharacharya (1025 CE), who gave the general formula, and by the Arab mathematician al-Khwarizmi (about 800 CE).

An equation may not look quadratic at first and must be simplified to standard form to decide. Take (x + 1)² = 2(x − 3). Expanding, x² + 2x + 1 = 2x − 6, so x² + 7 = 0, which is quadratic (a = 1, b = 0, c = 7). Take (x − 2)(x + 1) = (x − 1)(x + 3): expanding, x² − x − 2 = x² + 2x − 3, so −3x + 1 = 0, which is linear, not quadratic, because the x² terms cancel. Take x(2x + 3) = x² + 1: 2x² + 3x = x² + 1, so x² + 3x − 1 = 0, quadratic. Take (x + 2)³ = 2x(x² − 1): x³ + 6x² + 12x + 8 = 2x³ − 2x, so −x³ + 6x² + 14x + 8 = 0, which is cubic, not quadratic. The test is: after simplification, is the highest power exactly 2?

A real number α is a root or solution of the equation ax² + bx + c = 0 if aα² + bα + c = 0, that is, if substituting α makes the equation true. The roots of the equation are exactly the zeroes of the polynomial ax² + bx + c. Since a quadratic polynomial has at most two zeroes, a quadratic equation has at most two roots. Checking a root is substitution: is 2 a root of x² − 3x + 2 = 0? 4 − 6 + 2 = 0, yes. Is 3 a root of 2x² − 5x + 3 = 0? 18 − 15 + 3 = 6 ≠ 0, no. Is 3/2 a root? 2(9/4) − 15/2 + 3 = 9/2 − 15/2 + 6/2 = 0, yes.

In the examination the first questions ask you to check whether an equation is quadratic, to write it in standard form and name a, b, c, or to verify a given root. Simplify fully, collect all terms on one side, and write in descending powers before reading off the coefficients; a common slip is to read c from an unsimplified form.

Before solving quadratics we must be able to form them from words. The method is the same as for linear equations: name the unknown, express the other quantities in terms of it, and write the condition as an equation, which then turns out to have a square in it.

Areas. A rectangular plot has area 528 m² and its length is one more than twice its breadth. Let the breadth be x metres; the length is 2x + 1. Area = length × breadth: x(2x + 1) = 528, so 2x² + x − 528 = 0. This is the required quadratic.

Consecutive numbers. The product of two consecutive positive integers is 306. Let the smaller be x; the next is x + 1. Then x(x + 1) = 306, so x² + x − 306 = 0.

Ages. Rohan's mother is 26 years older than him. The product of their ages three years from now will be 360. Let Rohan's present age be x; his mother's is x + 26. In three years they will be x + 3 and x + 29, and (x + 3)(x + 29) = 360, so x² + 32x + 87 = 360, that is x² + 32x − 273 = 0.

Speed and time. A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Let the speed be x km/h; the time is 480/x hours. At x − 8 km/h the time is 480/(x − 8), which is 3 hours more: 480/(x − 8) − 480/x = 3. Multiply through by x(x − 8): 480x − 480(x − 8) = 3x(x − 8), so 3840 = 3x² − 24x, so 3x² − 24x − 3840 = 0, or dividing by 3, x² − 8x − 1280 = 0.

Cloth and cost. A shopkeeper buys a number of books for Rs 80. If he had bought 4 more books for the same amount, each would have cost Re 1 less. Let the number be x; the price each is 80/x. Then 80/(x + 4) = 80/x − 1; multiply by x(x + 4): 80x = 80(x + 4) − x(x + 4), so 80x = 80x + 320 − x² − 4x, so x² + 4x − 320 = 0.

Geometry. The hypotenuse of a right triangle is 13 cm and one side is 7 cm longer than the other. Let the shorter side be x; the other is x + 7. By Pythagoras, x² + (x + 7)² = 169, so 2x² + 14x + 49 = 169, 2x² + 14x − 120 = 0, x² + 7x − 60 = 0.

In each case the equation is written but not yet solved; forming it correctly is usually worth half the marks of a word problem. Keep the unknown clearly defined with its unit, expand carefully, and bring the equation to standard form with integer coefficients, dividing by a common factor if there is one. The next sections give the methods of solution, and the word problems return at the end with their answers.

If a product of two numbers is zero, at least one of them is zero. This zero-product rule is the basis of solving by factorisation: write the quadratic ax² + bx + c as a product of two linear factors, set each factor to zero, and solve the two linear equations. The roots of the quadratic equation are the zeroes of the two factors.

Worked example 1. Solve 2x² − 5x + 3 = 0. Split the middle term: we need two numbers whose product is 2 × 3 = 6 and whose sum is −5; they are −2 and −3. So 2x² − 2x − 3x + 3 = 0, that is 2x(x − 1) − 3(x − 1) = 0, that is (2x − 3)(x − 1) = 0. Hence 2x − 3 = 0 or x − 1 = 0, giving x = 3/2 or x = 1. Check: 2(9/4) − 15/2 + 3 = 0 and 2 − 5 + 3 = 0.

Worked example 2. Solve 6x² − x − 2 = 0. Product 6 × (−2) = −12, sum −1: the numbers are −4 and 3. 6x² − 4x + 3x − 2 = 2x(3x − 2) + 1(3x − 2) = (2x + 1)(3x − 2) = 0. Roots x = −1/2 and x = 2/3.

Worked example 3. Solve 3x² − 2√6 x + 2 = 0. Product 3 × 2 = 6, sum −2√6: the numbers are −√6 and −√6 (since √6 × √6 = 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), a repeated root.

Worked example 4. Solve 100x² − 20x + 1 = 0. This is (10x − 1)² = 0, so x = 1/10, repeated.

Worked example 5. Solve x² − 3x − 10 = 0: numbers −5 and 2; (x − 5)(x + 2) = 0; x = 5 or −2. Solve √2 x² + 7x + 5√2 = 0: product √2 × 5√2 = 10, sum 7; numbers 5 and 2; √2 x² + 5x + 2x + 5√2 = x(√2 x + 5) + √2(√2 x + 5) = (x + √2)(√2 x + 5) = 0; x = −√2 or −5/√2.

Word problem by factorisation. Find two numbers whose sum is 27 and product 182. Let one be x; the other is 27 − x. x(27 − x) = 182, so x² − 27x + 182 = 0. Numbers with product 182 and sum −27: −13 and −14. (x − 13)(x − 14) = 0, x = 13 or 14. The numbers are 13 and 14. Another: 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, x = 12 (reject −5). Base 12 cm, altitude 5 cm.

Factorisation is the quickest method when the roots are rational, which is the case in most textbook problems. It fails when the numbers cannot be found, which means the roots are irrational or not real; then use completing the square or the formula. Always check the roots by substitution and always reject a root that the problem forbids, stating the reason.

Some quadratics do not factorise with rational numbers, but every quadratic can be solved by completing the square: rewriting the equation so that the x-terms form a perfect square (x + k)², after which taking a square root gives x. The idea rests on the identity x² + 2kx + k² = (x + k)²: to complete the square on x² + 2kx we add k², which is the square of half the coefficient of x.

Steps for ax² + bx + c = 0. (1) Divide by a so the coefficient of x² is 1: x² + (b/a)x + c/a = 0. (2) Move the constant to the right: x² + (b/a)x = −c/a. (3) Add the square of half the coefficient of x, (b/2a)², to both sides. (4) Write the left side as (x + b/2a)². (5) Take square roots of both sides, remembering ±. (6) Solve for x.

Worked example 1. Solve x² + 4x − 5 = 0. 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.

Worked example 2. Solve 2x² − 7x + 3 = 0. Divide by 2: x² − (7/2)x + 3/2 = 0, so 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 + 49)/16 = 25/16. So (x − 7/4)² = 25/16, x − 7/4 = ±5/4, x = 7/4 + 5/4 = 3 or x = 7/4 − 5/4 = 1/2. Roots 3 and 1/2.

Worked example 3. Solve 2x² + x − 4 = 0. Divide by 2: x² + x/2 = 2. Add (1/4)² = 1/16: x² + x/2 + 1/16 = 2 + 1/16 = 33/16. (x + 1/4)² = 33/16, x + 1/4 = ±√33/4, x = (−1 ± √33)/4. These irrational roots could not have been found by factorisation.

Worked example 4. Solve 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: x² + √3 x + 3/4 = 0, so (x + √3/2)² = 0, x = −√3/2, a repeated root.

Worked example 5 (no real root). Solve 2x² + x + 4 = 0. Divide by 2: x² + x/2 = −2. Add 1/16: (x + 1/4)² = −2 + 1/16 = −31/16. The square of a real number cannot be negative, so there is no real root. Completing the square thus also detects when an equation has no real solution.

A variant avoids fractions: multiply the equation by 4a instead of dividing by a. For 2x² − 7x + 3 = 0, multiply by 8: 16x² − 56x + 24 = 0, so (4x)² − 2(4x)(7) + 49 = 25, (4x − 7)² = 25, 4x − 7 = ±5, x = 3 or 1/2. Either way, the method is longer than factorisation but never fails, and it is the method by which the quadratic formula is derived in the next section.

Completing the square on the general equation once and for all gives a formula that solves every quadratic. 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 to both sides: x² + (b/a)x + b²/4a² = b²/4a² − c/a = (b² − 4ac)/4a². The left side is a perfect square: (x + b/2a)² = (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, known in India as Sridharacharya's formula. The two signs give the two roots: α = [−b + √(b² − 4ac)]/2a and β = [−b − √(b² − 4ac)]/2a. If b² − 4ac < 0 there is no real root, because a negative number has no real square root.

Worked example 1. Solve 2x² − 7x + 3 = 0. a = 2, b = −7, c = 3. b² − 4ac = 49 − 24 = 25. x = [7 ± 5]/4, so x = 12/4 = 3 or x = 2/4 = 1/2. Same as by completing the square.

Worked example 2. Solve x² + 4x + 5 = 0. b² − 4ac = 16 − 20 = −4 < 0. No real roots.

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

Worked example 4. Solve x² − 3x − 1 = 0 (irrational roots). b² − 4ac = 9 + 4 = 13. x = (3 ± √13)/2. Both are real; approximately 3.30 and −0.30.

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

Worked example 6 (reciprocal equation). Solve 1/x − 1/(x − 2) = 3, x ≠ 0, 2. Combine: [(x − 2) − x]/[x(x − 2)] = 3, so −2 = 3x² − 6x, so 3x² − 6x + 2 = 0. b² − 4ac = 36 − 24 = 12. x = [6 ± √12]/6 = [6 ± 2√3]/6 = 1 ± √3/3 = (3 ± √3)/3. Both allowed since neither is 0 or 2.

Worked example 7. Solve x + 1/x = 3, x ≠ 0. Multiply by x: x² − 3x + 1 = 0. x = (3 ± √5)/2.

The formula is the universal method: it works for every quadratic, gives irrational roots exactly, and tells you at once from the sign of b² − 4ac whether real roots exist. Its only danger is arithmetic: write down a, b, c with their signs, compute b² − 4ac separately, and simplify the surd before writing the final answer. Where the roots are rational, factorisation is faster and is the expected method unless the question names another.

In the quadratic formula everything under the square root is the quantity b² − 4ac, called the discriminant and denoted D (or Δ). It discriminates, that is decides, what kind of roots the equation has, without our solving it.

Case 1: D > 0. √D is a positive real number, so the ± gives two different values: the equation has two distinct real roots, [−b + √D]/2a and [−b − √D]/2a. If D is a perfect square (like 25 or 1) the roots are rational and the equation factorises; if D is not a perfect square (like 13) the roots are irrational and come as a pair p ± q√r.

Case 2: D = 0. √D = 0, so both signs give the same value: the equation has two equal real roots, each equal to −b/2a. The quadratic is then a perfect square, a(x + b/2a)², and its graph touches the x-axis at one point.

Case 3: D < 0. √D is not a real number, so the equation has no real roots. The graph of y = ax² + bx + c lies wholly above or wholly below the x-axis.

Worked examples. 2x² − 4x + 3 = 0: D = 16 − 24 = −8 < 0, no real roots. 2x² − 6x + 3 = 0: D = 36 − 24 = 12 > 0, two distinct real roots, x = (6 ± √12)/4 = (3 ± √3)/2. 3x² − 4√3 x + 4 = 0: D = 48 − 48 = 0, equal roots, x = 4√3/6 = 2/√3 each. x² + 4x + 4 = 0: D = 0, x = −2 twice. 2x² + x − 4 = 0: D = 1 + 32 = 33, two distinct irrational roots.

Finding k for equal roots. If the question says an equation has equal roots, set D = 0 and solve for the unknown. For 2x² + kx + 3 = 0: D = k² − 24 = 0, k = ±2√6. For kx(x − 2) + 6 = 0, that is kx² − 2kx + 6 = 0: D = 4k² − 24k = 0, 4k(k − 6) = 0, so k = 6 (k = 0 is rejected because then the equation is not quadratic). For x² − 2(k + 1)x + k² = 0: D = 4(k + 1)² − 4k² = 4(2k + 1) = 0, k = −1/2. For (k + 1)x² − 2(k − 1)x + 1 = 0: D = 4(k − 1)² − 4(k + 1) = 4(k² − 3k) = 0, k = 0 or 3.

Finding k for real roots. For real roots we need D ≥ 0. For x² + kx + 4 = 0: k² − 16 ≥ 0, so k ≥ 4 or k ≤ −4. For 2x² + 3x + k = 0: 9 − 8k ≥ 0, k ≤ 9/8. For no real roots we need D < 0: x² + 5x + k = 0 has no real roots when 25 − 4k < 0, k > 25/4.

Word problems using D. Is it possible to design a rectangular mango grove whose length is twice its breadth and area 800 m²? Breadth x: 2x² = 800, x² = 400, x = 20; yes, 20 m by 40 m. Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? Length x, breadth 40 − x: x(40 − x) = 400, x² − 40x + 400 = 0, D = 1600 − 1600 = 0, x = 20; yes, a square of side 20 m. Two friends' ages total 20 years; four years ago the product of their ages was 48: (x − 4)(16 − x) = 48 gives x² − 20x + 112 = 0, D = 400 − 448 < 0, so no such ages exist; the situation is impossible.

The discriminant is the single most examined idea of the chapter. Memorise: D > 0 two distinct real roots, D = 0 two equal real roots, D < 0 no real roots; and use it both to describe roots and to find unknown coefficients.

With the methods in hand, the word problems set up earlier can be finished. The pattern for every word problem: define the unknown with units; form the equation; write it in standard form; solve by the most suitable method; check both roots against the conditions of the problem, rejecting any that are impossible (negative lengths, fractional numbers of people, ages that make someone unborn); and answer in words.

Two consecutive positive integers with product 306. x² + x − 306 = 0. Product −306, sum 1: numbers 18 and −17. (x + 18)(x − 17) = 0, x = 17 or −18. Reject −18 (not positive). The integers are 17 and 18.

Rohan's age. x² + 32x − 273 = 0. Product −273, sum 32: numbers 39 and −7. (x + 39)(x − 7) = 0, x = 7 (reject −39). Rohan is 7 and his mother is 33. Check: in three years, 10 × 36 = 360.

Sum and squares. 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 − 286 = 0, x² + 2x − 143 = 0, (x + 13)(x − 11) = 0, x = 11. The integers are 11 and 13. Check: 121 + 169 = 290.

A number and its reciprocal. 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 are valid; each is the reciprocal of the other.

Difference of squares. The difference of squares of two numbers is 180 and the square of the smaller is 8 times the larger. Larger x, smaller y: y² = 8x and x² − y² = 180, so x² − 8x − 180 = 0, (x − 18)(x + 10) = 0, x = 18 (reject −10 since then y² = −80). y² = 144, y = ±12. The numbers are 18 and 12, or 18 and −12.

Ages with a product. One year ago a man was 8 times as old as his son; now his age equals the square of his son's age. Son's present age x, man's x²: x² − 1 = 8(x − 1), so x² − 8x + 7 = 0, (x − 1)(x − 7) = 0, x = 7 (reject x = 1, since then the man would be 1 year old). The son is 7 and the man 49. Check: a year ago 48 = 8 × 6.

Rehman. The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. 1/(x − 3) + 1/(x + 5) = 1/3. Combine: (x + 5 + x − 3)/[(x − 3)(x + 5)] = 1/3, so 3(2x + 2) = x² + 2x − 15, so x² − 4x − 21 = 0, (x − 7)(x + 3) = 0, x = 7 (reject −3). Rehman is 7 years old. Check: 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3.

Marks. In a class test the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 less in English, the product would have been 210. Mathematics x, English 30 − x: (x + 2)(27 − x) = 210, so −x² + 25x + 54 = 210, x² − 25x + 156 = 0, (x − 12)(x − 13) = 0. Two answers: Mathematics 12 and English 18, or Mathematics 13 and English 17. Both satisfy the problem and both must be stated.

Problems on motion and work produce quadratics through the reciprocal relation between speed and time, or between rate and time. The setting up always uses time = distance/speed or work rate = 1/(time taken alone).

The train of 480 km. We found x² − 8x − 1280 = 0 for the speed x. Factorise: product −1280, sum −8: numbers −40 and 32. (x − 40)(x + 32) = 0, x = 40 (reject −32). The speed is 40 km/h. Check: at 40 km/h the time is 12 h; at 32 km/h it is 15 h, 3 hours more.

Two trains. 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 is 11 km/h more. Let the passenger train's speed be x km/h: 132/x − 132/(x + 11) = 1. Multiply by x(x + 11): 132(x + 11) − 132x = x² + 11x, so 1452 = x² + 11x, x² + 11x − 1452 = 0. Product −1452, sum 11: 44 and −33. (x + 44)(x − 33) = 0, x = 33. Passenger 33 km/h, express 44 km/h. Check: 132/33 = 4 h, 132/44 = 3 h.

Motor boat. 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. Let the stream's speed be x: 24/(18 − x) − 24/(18 + x) = 1. Multiply by (18 − x)(18 + x) = 324 − x²: 24(18 + x) − 24(18 − x) = 324 − x², so 48x = 324 − x², x² + 48x − 324 = 0. Product −324, sum 48: 54 and −6. (x + 54)(x − 6) = 0, x = 6. The stream flows at 6 km/h. Check: upstream 24/12 = 2 h, downstream 24/24 = 1 h.

Aeroplane. A plane left 30 minutes late and, to reach 1500 km on time, increased its speed by 100 km/h. Let the usual speed be x: 1500/x − 1500/(x + 100) = 1/2. Multiply by 2x(x + 100): 3000(x + 100) − 3000x = x² + 100x, so 300000 = x² + 100x, x² + 100x − 300000 = 0, (x + 600)(x − 500) = 0, x = 500 km/h.

Two taps. Two water taps together fill a tank in 9⅜ = 75/8 hours. The tap of larger diameter takes 10 hours less than the smaller one alone. Let the smaller tap take x hours; the larger takes x − 10. Rates: 1/x + 1/(x − 10) = 8/75. Combine: (2x − 10)/[x(x − 10)] = 8/75, so 75(2x − 10) = 8(x² − 10x), 150x − 750 = 8x² − 80x, 8x² − 230x + 750 = 0, 4x² − 115x + 375 = 0. Product 1500, sum −115: −100 and −15. 4x² − 100x − 15x + 375 = 4x(x − 25) − 15(x − 25) = (4x − 15)(x − 25) = 0, x = 25 or 15/4. Reject 15/4 because then x − 10 is negative. Smaller tap 25 h, larger tap 15 h. Check: 1/25 + 1/15 = (3 + 5)/75 = 8/75.

Two pipes. A cistern is filled by two pipes in 6 minutes together; alone, one takes 5 minutes more than the other. Faster pipe x: 1/x + 1/(x + 5) = 1/6, 6(2x + 5) = x² + 5x, x² − 7x − 30 = 0, (x − 10)(x + 3) = 0, x = 10. Pipes take 10 and 15 minutes.

The equations here are long to clear of fractions; multiply by the product of the denominators, expand carefully, and simplify by a common factor before factorising. The extra root is almost always negative or makes a time negative, and must be rejected with a stated reason.

The rectangular plot. 2x² + x − 528 = 0 for the breadth. Product −1056, sum 1: 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.

The right triangle. x² + 7x − 60 = 0 for the shorter side. (x + 12)(x − 5) = 0, x = 5. Sides 5 cm, 12 cm, hypotenuse 13 cm.

Two squares. The sum of the areas of two squares is 468 m² and the difference of their perimeters is 24 m. Sides x and y with x > y: 4x − 4y = 24 gives x = y + 6; x² + y² = 468 gives (y + 6)² + y² = 468, 2y² + 12y − 432 = 0, y² + 6y − 216 = 0, (y + 18)(y − 12) = 0, y = 12, x = 18. Sides 18 m and 12 m.

A 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 AB is a diameter, the angle at the pole is a right angle. Distances x and x + 7: x² + (x + 7)² = 169, x² + 7x − 60 = 0, x = 5. The pole is 5 m from one gate and 12 m from the other; it is possible.

Two standard problems on cost of production follow the same pattern.

Cottage industry. A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article was 3 more than twice the number of articles produced, and the total cost was Rs 90. Let the number be x; cost each is 2x + 3; x(2x + 3) = 90, 2x² + 3x − 90 = 0, product −180, sum 3: 15 and −12; 2x² + 15x − 12x − 90 = x(2x + 15) − 6(2x + 15) = (2x + 15)(x − 6) = 0, x = 6. Six articles at Rs 15 each.

Toys. On a particular day the cost of production of each toy was 55 minus the number of toys produced, and the total cost was Rs 750. x(55 − x) = 750, x² − 55x + 750 = 0, (x − 25)(x − 30) = 0, x = 25 or 30. Both are possible: 25 toys at Rs 30 or 30 toys at Rs 25.

Handshakes and matches. In a group of x people every pair shakes hands once; the number of handshakes is x(x − 1)/2. If there were 66 handshakes, x² − x − 132 = 0, (x − 12)(x + 11) = 0, x = 12 people. Similarly, in a league where every team plays every other twice, x(x − 1) matches are played; 132 matches gives x = 12 teams.

Frame around a picture. A picture 24 cm by 18 cm has a frame of uniform width x around it, and the area of the frame equals the area of the picture. (24 + 2x)(18 + 2x) − 432 = 432, so 4x² + 84x + 432 − 864 = 0, x² + 21x − 108 = 0, product −108, sum 21 gives 24 and −3, so (x + 24)(x − 3) = 0, x = 3 cm (reject −24).

These problems reward drawing a small figure, labelling the unknown on it, and writing the geometric relation (area, Pythagoras, perimeter) before any algebra. As always, check the root against the figure: a negative or a too-large width is rejected.

Because the roots of ax² + bx + c = 0 are the zeroes of the polynomial ax² + bx + c, the relations from the previous chapter hold: if α and β are the roots then α + β = −b/a and αβ = c/a. They can also be read from the formula: α + β = [(−b + √D) + (−b − √D)]/2a = −2b/2a = −b/a, and αβ = [(−b)² − (√D)²]/4a² = (b² − b² + 4ac)/4a² = c/a. And the difference: α − β = √D/a (taking α as the larger).

Forming an equation from its roots. An equation with roots α and β is x² − (α + β)x + αβ = 0, or any non-zero multiple. Roots 3 and −5: x² + 2x − 15 = 0. Roots 2 + √3 and 2 − √3: sum 4, product 1: x² − 4x + 1 = 0. Roots 1/2 and −1/3: sum 1/6, product −1/6: x² − x/6 − 1/6 = 0, or 6x² − x − 1 = 0.

Using the relations without solving. If α, β are roots of 2x² − 5x + 3 = 0, find α² + β² and 1/α + 1/β. α + β = 5/2, αβ = 3/2. α² + β² = (α + β)² − 2αβ = 25/4 − 3 = 13/4. 1/α + 1/β = (α + β)/αβ = (5/2)/(3/2) = 5/3. Find α³ + β³: (α + β)³ − 3αβ(α + β) = 125/8 − (9/2)(5/2) = 125/8 − 45/4 = 125/8 − 90/8 = 35/8.

Conditions on roots. If one root of x² − 6x + k = 0 is twice the other: roots α and 2α, sum 3α = 6, α = 2, product 2α² = 8 = k. If the roots of x² + px + 12 = 0 differ by 1: (α − β)² = (α + β)² − 4αβ = p² − 48 = 1, p² = 49, p = ±7. If one root of 3x² − 10x + k = 0 is the reciprocal of the other: product k/3 = 1, k = 3. If the roots of 2x² + kx + 8 = 0 are equal: D = k² − 64 = 0, k = ±8 (or by the relations, α = β means (α + β)² = 4αβ, k²/4 = 16).

An equation whose roots are related to another's. If α, β are roots of x² − 3x + 2 = 0, form the equation whose roots are α + 1 and β + 1. New sum = (α + β) + 2 = 5, new product = αβ + (α + β) + 1 = 2 + 3 + 1 = 6. Equation: x² − 5x + 6 = 0. Check: the original roots are 1 and 2, so the new roots are 2 and 3, and (x − 2)(x − 3) = x² − 5x + 6.

Sign of the roots. Without solving, the signs of the roots can be read from the relations: if the product c/a is negative the roots have opposite signs; if the product is positive and the sum −b/a is positive both roots are positive; if the product is positive and the sum negative both are negative. For x² − 5x + 6 = 0, product 6 > 0 and sum 5 > 0, so both roots positive (2 and 3). For x² + 5x + 6 = 0, both negative (−2, −3). For x² − x − 6 = 0, opposite signs (3, −2).

These relations let many questions be answered in two lines instead of solving and substituting; they are also the check for every solved equation: add and multiply the roots you found and compare with −b/a and c/a.

The graph. The real roots of ax² + bx + c = 0 are the x-coordinates of the points where the parabola y = ax² + bx + c meets the x-axis. So the three discriminant cases are three pictures: D > 0, the parabola cuts the axis at two points; D = 0, it touches the axis at its vertex; D < 0, it stays entirely on one side. The vertex is at x = −b/2a, which is also the average of the two roots, (α + β)/2, and the axis of symmetry passes through it. For y = x² − 5x + 6 the roots are 2 and 3, the vertex is at x = 2.5 with y = 6.25 − 12.5 + 6 = −0.25, just below the axis, which is why the two roots are so close together; for y = x² − 5x + 6.25 the vertex would sit on the axis and the roots would coincide at 2.5; for y = x² − 5x + 7 the vertex is above the axis and there are no real roots.

Graphs also solve equations approximately: draw y = ax² + bx + c on graph paper and read where it crosses. For x² − 3x − 1 = 0, the parabola crosses near x = 3.3 and x = −0.3, matching (3 ± √13)/2.

Equations reducible to quadratics. Some equations of higher degree or with fractions become quadratic by a substitution. Biquadratic: x⁴ − 5x² + 4 = 0. Put y = x²: y² − 5y + 4 = 0, (y − 1)(y − 4) = 0, y = 1 or 4, so x² = 1 or x² = 4, x = ±1, ±2. Four roots. Reciprocal: x + 1/x = 5/2. Multiply by 2x: 2x² − 5x + 2 = 0, (2x − 1)(x − 2) = 0, x = 2 or 1/2. Rational: (x − 1)/(x + 1) + (x + 1)/(x − 1) = 5/2, x ≠ ±1. Put t = (x − 1)/(x + 1): t + 1/t = 5/2, 2t² − 5t + 2 = 0, t = 2 or 1/2. If (x − 1)/(x + 1) = 2 then x − 1 = 2x + 2, x = −3; if (x − 1)/(x + 1) = 1/2 then 2x − 2 = x + 1, x = 3. Roots ±3. Radical: √(x + 5) = x − 1. Square: x + 5 = x² − 2x + 1, x² − 3x − 4 = 0, (x − 4)(x + 1) = 0, x = 4 or −1. Check in the original (squaring can introduce false roots): x = 4 gives √9 = 3 = 4 − 1, valid; x = −1 gives √4 = 2 but −1 − 1 = −2, invalid. Only x = 4.

Equations with a common factor. x³ − 4x = 0 is x(x² − 4) = 0, giving x = 0, ±2; a cubic solved through a quadratic factor. (x² − 5x)² − 2(x² − 5x) − 24 = 0: put y = x² − 5x, y² − 2y − 24 = (y − 6)(y + 4) = 0, y = 6 or −4; x² − 5x − 6 = 0 gives x = 6, −1; x² − 5x + 4 = 0 gives x = 4, 1. Four roots: −1, 1, 4, 6.

In every substitution problem, solve for the new variable, return to x, and check each answer in the original equation, especially after squaring or after clearing a denominator that might be zero. The substitution technique is the bridge to the higher-degree equations of the Intermediate course.

Quadratic equations carry a large share of the algebra marks in the Class 10 paper. The question types and the traps are as follows.

One-mark. Is a given equation quadratic? Write it in standard form. State the discriminant of a given equation. State the nature of roots when D is given. Find k for equal roots in a simple equation such as x² + kx + 9 = 0 (k = ±6). Give the roots of x² − 4 = 0 (±2) or x² = 3 (±√3). State the sum and product of the roots of a given equation.

Two-mark. Solve by factorisation. Find the discriminant and hence the nature of the roots. Find k for equal roots with a little algebra. Check whether a given number is a root. Form the equation with given roots.

Four-mark. Solve by completing the square or by the formula, with irrational roots left exact. Solve a reciprocal or rational equation. A word problem on numbers, ages or areas. Find k for real roots (an inequality) or for no real roots.

Eight-mark. A long word problem on speed and time, or on taps and pipes, with formulation, solution, rejection of the inadmissible root and verification; or a proof-type question such as showing that a given equation has real roots for all k (show D ≥ 0 as a perfect square or a sum of squares), for example, for x² + 2(k + 1)x + 2k = 0, D = 4(k + 1)² − 8k = 4(k² + 1) > 0 for all k, so the roots are always real and distinct.

Common errors. (1) Not reducing to standard form before reading a, b, c, especially the sign of c. (2) Dropping the ± when taking a square root, giving only one root. (3) Writing the formula with 2a under −b only, instead of under the whole numerator. (4) Arithmetic in D: b² is always non-negative; −4ac takes the sign of −ac. (5) Forgetting to divide by a before completing the square. (6) Accepting a negative length, speed or age; or, the opposite error, rejecting a negative root in a pure number problem where it is valid. (7) In equations with denominators, not stating the excluded values (x ≠ 0, x ≠ 2) and not rejecting a root that makes a denominator zero. (8) In 'equal roots' problems, keeping k = 0 when it makes the equation non-quadratic. (9) After squaring an equation, not checking for extraneous roots. (10) Forgetting that a quadratic may have two valid answers to a word problem (the toys, Shefali's marks) and giving only one.

Summary. ax² + bx + c = 0, a ≠ 0. Solve by factorisation (zero-product rule), by completing the square, or by x = [−b ± √(b² − 4ac)]/2a. D = b² − 4ac: positive gives two distinct real roots, zero gives two equal roots −b/2a, negative gives no real roots. α + β = −b/a, αβ = c/a. Word problems: define, formulate, solve, reject, verify, answer in words. Master these and the chapter is yours.