1. Chapter at a Glance
A rectangular prayer hall needs an area of 300 m², with length 1 m more than twice its breadth. What are its dimensions? Writing b for the breadth gives b(2b + 1) = 300, i.e. 2b² + b − 300 = 0 — a quadratic equation. Indian mathematicians studied such equations long ago: Brahmagupta (7th century) gave a rule for solving them, and Śrīdharācārya (around 1025 CE) gave the method that leads to the quadratic formula. In this chapter you will solve quadratic equations by factorisation and by the quadratic formula, and use the discriminant to predict the nature of roots.
- Standard form ax² + bx + c = 0 (a ≠ 0) and checking if an equation is quadratic.
- Roots of a quadratic equation; checking a root.
- Solving by factorisation (splitting the middle term).
- The quadratic formula.
- Discriminant D = b² − 4ac and the nature of roots.
- Word problems: area, numbers, speed, age, work.
2. Standard Form
A quadratic equation in x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
Check: (a) (x + 2)³ = x³ − 4 (b) (x − 2)² + 1 = 2x − 3.
Solution: (a) x³ + 6x² + 12x + 8 = x³ − 4 → 6x² + 12x + 12 = 0 → yes (x³ cancels). (b) x² − 4x + 5 = 2x − 3 → x² − 6x + 8 = 0 → yes.
A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. The roots of the equation are the zeros of the polynomial ax² + bx + c. A quadratic equation has at most two roots.
3. Solving by Factorisation
- Write the equation in standard form.
- Split the middle term: find two numbers whose product = ac and sum = b.
- Factorise into two linear factors.
- Set each factor to zero (if pq = 0 then p = 0 or q = 0).
Solve 2x² − 5x + 3 = 0.
Solution: ac = 6; −2 and −3. 2x² − 2x − 3x + 3 = 2x(x − 1) − 3(x − 1) = (2x − 3)(x − 1) = 0 → x = 3/2 or x = 1.
Solve √2x² + 7x + 5√2 = 0.
Solution: ac = 10; 5 and 2. √2x² + 5x + 2x + 5√2 = x(√2x + 5) + √2(√2x + 5) = (x + √2)(√2x + 5) → x = −√2 or x = −5/√2 = −5√2/2.
4. The Quadratic Formula
The roots of ax² + bx + c = 0 are
Where it comes from: multiply ax² + bx + c = 0 by 4a: 4a²x² + 4abx = −4ac. Add b² to both sides: (2ax + b)² = b² − 4ac. Take square roots: 2ax + b = ±√(b² − 4ac), giving the formula.
Solve 3x² − 5x + 2 = 0 using the formula.
Solution: D = 25 − 24 = 1. x = (5 ± 1)/6 → x = 1 or x = 2/3.
Solve x² − 4x + 1 = 0.
Solution: D = 16 − 4 = 12. x = (4 ± 2√3)/2 = 2 ± √3.
5. Nature of Roots — The Discriminant
The quantity D = b² − 4ac is called the discriminant; it "discriminates" between the kinds of roots.
| Discriminant | Nature of roots | Graph of y = ax² + bx + c |
|---|---|---|
| D > 0 | Two distinct real roots | Cuts the x-axis at two points |
| D = 0 | Two equal real roots, x = −b/2a | Touches the x-axis at one point |
| D < 0 | No real roots | Does not meet the x-axis |
Find k so that kx(x − 2) + 6 = 0 has two equal roots.
Solution: kx² − 2kx + 6 = 0. D = 4k² − 24k = 0 → 4k(k − 6) = 0 → k = 6 (k = 0 makes it not quadratic). k = 6.
6. Word Problems
Solve 2b² + b − 300 = 0 from the introduction.
Solution: ac = −600; 25 and −24. 2b² + 25b − 24b − 300 = b(2b + 25) − 12(2b + 25) = (b − 12)(2b + 25). b = 12 (breadth cannot be negative). Breadth 12 m, length 25 m.
A train travels 360 km at a uniform speed. If the speed were 5 km/h more, it would take 1 hour less. Find the speed.
Solution: 360/x − 360/(x + 5) = 1 → 1800 = x² + 5x → x² + 5x − 1800 = 0 → (x + 45)(x − 40) = 0 → 40 km/h.
7. Solved Problems (Board Pattern)
Find two consecutive positive integers whose squares sum to 365.
Solution: x² + (x + 1)² = 365 → 2x² + 2x − 364 = 0 → x² + x − 182 = 0 → (x + 14)(x − 13) = 0 → 13 and 14.
The altitude of a right triangle is 7 cm less than its base, and the hypotenuse is 13 cm. Find the other two sides.
Solution: x² + (x − 7)² = 169 → 2x² − 14x − 120 = 0 → x² − 7x − 60 = 0 → (x − 12)(x + 5) = 0 → base 12 cm, altitude 5 cm.
Find the nature of roots of 2x² − 6x + 3 = 0.
Solution: D = 36 − 24 = 12 > 0 → two distinct real roots ((3 ± √3)/2).
8. Master Formula Table
| Result | Formula |
|---|---|
| Standard form | ax² + bx + c = 0, a ≠ 0 |
| Quadratic formula | x = [−b ± √(b² − 4ac)] / 2a |
| Discriminant | D = b² − 4ac |
| Equal roots | x = −b/2a (when D = 0) |
| Sum of roots | −b/a |
| Product of roots | c/a |
| Splitting middle term | p + q = b, pq = ac |
9. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Recognise quadratic | Simplify; check x² term remains | MCQ / 1 mark |
| Factorisation | Split middle term | Solve quickly |
| Formula | Works always (D ≥ 0) | Irrational roots |
| Discriminant | Sign of D → nature | Find k for equal roots |
| Word problems | Reject impossible roots | Area, speed, age, numbers |
10. Mind Map
- ax² + bx + c = 0
- a ≠ 0
- Factorise
- Formula
- >0, =0, <0
- Equal roots −b/2a
- Area, speed
- Numbers, ages
11. Tips and Tricks
- Divide out common factors first (2x² + 2x − 364 → x² + x − 182).
- Compute D first: if it is a perfect square, the equation factorises nicely.
- In speed/time problems, the equation usually has the form 360/x − 360/(x + 5) = 1 — clear the denominators carefully.
- Dividing both sides by x and losing the root x = 0.
- Keeping a negative length, speed or age in word problems.
- Sign error in −b when b is negative.
12. Important One-liners
- A quadratic equation has at most two roots.
- D > 0: two distinct real roots; D = 0: two equal roots; D < 0: no real roots.
- If a and c have opposite signs, D > 0 always (real roots).
- Equal roots are each −b/2a.
- Śrīdharācārya gave the method behind the quadratic formula.
13. Exam-pattern Questions with Detailed Answers
Which is a quadratic equation?
- x² + 2x + 1 = (4 − x)² + 3
- x(x + 1) + 8 = (x + 2)(x − 2)
- x(2x + 3) = x² + 1
- (x + 2)² = x² + 4
Show answer
(C): 2x² + 3x = x² + 1 → x² + 3x − 1 = 0. The others lose the x² term.
Answer: (C)
Find the discriminant of 2x² − 4x + 3 = 0 and the nature of its roots.
Show answer
D = 16 − 24 = −8 < 0 → no real roots.
Answer: D = −8; no real roots
Solve x² − 3x − 10 = 0.
Show answer
(x − 5)(x + 2) = 0 → 5, −2.
Answer: x = 5 or −2
Solve 2x² + x − 4 = 0 using the quadratic formula.
Show answer
D = 1 + 32 = 33. x = (−1 ± √33)/4.
Answer: (−1 ± √33)/4
Find k if 2x² + kx + 3 = 0 has two equal roots.
Show answer
k² − 24 = 0 → k = ±2√6.
Answer: k = ±2√6
Solve 1/x − 1/(x − 2) = 3, x ≠ 0, 2.
Show answer
(x − 2 − x)/(x(x − 2)) = 3 → −2 = 3x² − 6x → 3x² − 6x + 2 = 0. x = (6 ± √(36 − 24))/6 = (3 ± √3)/3.
Answer: x = (3 ± √3)/3
The product of two consecutive positive integers is 306. Find them.
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x² + x − 306 = 0 → (x + 18)(x − 17) = 0 → 17, 18.
Answer: 17 and 18
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
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l + b = 40, lb = 400 → l(40 − l) = 400 → l² − 40l + 400 = 0 → (l − 20)² = 0 → l = 20. Yes: a 20 m × 20 m square park.
Answer: Yes; 20 m × 20 m
A motor boat 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.
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24/(18 − x) − 24/(18 + x) = 1 → 48x = 324 − x² → x² + 48x − 324 = 0 → (x + 54)(x − 6) = 0 → 6 km/h.
Answer: 6 km/h
Two water taps together fill a tank in 75/8 hours. The larger tap takes 10 hours less than the smaller one to fill it alone. Find the time each takes alone.
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1/x + 1/(x − 10) = 8/75 → 75(2x − 10) = 8x(x − 10) → 8x² − 230x + 750 = 0 → 4x² − 115x + 375 = 0 → (x − 25)(4x − 15) = 0. x = 25 (x = 3.75 makes x − 10 negative). Smaller tap 25 h, larger 15 h.
Answer: 25 hours and 15 hours
The sum of the reciprocals of Rehman's ages 3 years ago and 5 years from now is 1/3. Find his present age.
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1/(x − 3) + 1/(x + 5) = 1/3 → 3(2x + 2) = (x − 3)(x + 5) → 6x + 6 = x² + 2x − 15 → x² − 4x − 21 = 0 → (x − 7)(x + 3) = 0 → 7 years.
Answer: 7 years
Find the values of p for which (p + 1)x² − 6(p + 1)x + 3(p + 9) = 0 has equal roots (p ≠ −1). Then find the roots.
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D = 36(p + 1)² − 12(p + 1)(p + 9) = 12(p + 1)[3(p + 1) − (p + 9)] = 12(p + 1)(2p − 6) = 0 → p = 3. Equation: 4x² − 24x + 36 = 0 → (x − 3)² = 0 → roots 3, 3.
Answer: p = 3; roots 3, 3
14. Quick Revision Summary
- ax² + bx + c = 0, a ≠ 0; at most two roots.
- Factorise by splitting the middle term (product ac, sum b).
- Formula: x = [−b ± √(b² − 4ac)]/2a.
- D > 0 distinct, D = 0 equal (−b/2a), D < 0 no real roots.
- In word problems, reject roots that make no physical sense.