Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
A quadratic equation in x is an equation of the form
This is its standard form: terms written in descending powers of x. Any equation p(x) = 0, where p(x) is a polynomial of degree 2, is quadratic.
Always expand and collect terms first. Something that looks quadratic may not be, and something that looks cubic may be quadratic.
Choose x for the unknown, express the other quantities in x, and use the given condition.
A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. The roots of the equation are the zeroes of the polynomial ax² + bx + c, so there are at most two roots.
Check each root against the situation. A length, a number of people or a speed cannot be negative, so a negative root is rejected. For the prayer hall, 2x² + x − 300 = (x − 12)(2x + 25) = 0 gives x = 12 or −12.5; the breadth is 12 m and the length 25 m.
The roots of ax² + bx + c = 0 are
This works for every quadratic equation with real roots, even when factorising is hard.
The number D = b² − 4ac is the discriminant. It decides the nature of the roots:
| Discriminant | Nature of roots | Roots |
|---|---|---|
| D > 0 | two distinct real roots | −b + √D2a and −b − √D2a |
| D = 0 | two equal real roots | both equal to −b2a |
| D < 0 | no real roots | no real number has a negative square |
Many questions ask whether a situation is possible. Form the quadratic and check the discriminant: if D < 0, the situation is impossible. The pole example below is possible because D = 289 > 0.
Set D = 0 and solve for k. For 2x² + kx + 3 = 0: k² − 24 = 0, so k = ±2√6. Exclude any k that makes a = 0.
| Type | Let | Equation from |
|---|---|---|
| Two numbers, sum s | x and s − x | product given |
| Consecutive integers | x and x + 1 | product or sum of squares given |
| Consecutive odd/even | x and x + 2 | product given |
| Ages | present age x | product of ages n years ago/later |
| Rectangle | breadth x | area = length × breadth |
| Right triangle | one side x | Pythagoras' theorem |
| Speed | speed x km/h | time = distance ÷ speed; difference of times |
| Sharing a cost | x people | cost each = total ÷ x; difference of shares |
A train covers 480 km. If its speed were 8 km/h less it would take 3 hours more:
The speed is 40 km/h (the negative root is rejected).
| Item | Result |
|---|---|
| Standard form | ax² + bx + c = 0, a ≠ 0 |
| Quadratic formula | x = (−b ± √(b² − 4ac)) / 2a |
| Discriminant | D = b² − 4ac |
| D > 0 / D = 0 / D < 0 | two distinct / two equal / no real roots |
| Equal roots | each root = −b/2a |
Try each problem first, then open the solution to check your working.
Check whether each is a quadratic equation: (a) (x + 3)² = x² − 5 (b) x(x − 4) = 2x² + 1 (c) (2x − 1)³ = 8x³ + x
Use: Expand and collect all terms on one side
(a) x² + 6x + 9 = x² − 5 ⇒ 6x + 14 = 0: degree 1, not quadratic.
(b) x² − 4x = 2x² + 1 ⇒ x² + 4x + 1 = 0: quadratic.
(c) 8x³ − 12x² + 6x − 1 = 8x³ + x ⇒ 12x² − 5x + 1 = 0: quadratic.
Answer: (a) no (b) yes (c) yes
The sum of a number and its reciprocal is 103. Form a quadratic equation and find the number.
Given: number x (x ≠ 0)
Use: x + 1x = 103
Multiply by 3x: 3x² + 3 = 10x ⇒ 3x² − 10x + 3 = 0
3x² − 9x − x + 3 = (3x − 1)(x − 3) = 0
Answer: 3x² − 10x + 3 = 0; the number is 3 or 1/3
Solve x² + 5x − 24 = 0.
Use: Two numbers with product −24 and sum 5: 8 and −3
x² + 8x − 3x − 24 = (x + 8)(x − 3) = 0
Answer: x = −8 or x = 3
Solve 6x² + 7x − 3 = 0.
Given: ac = −18
Use: Split 7x as 9x − 2x
6x² + 9x − 2x − 3 = 3x(2x + 3) − 1(2x + 3) = (3x − 1)(2x + 3) = 0
Answer: x = 1/3 or x = −3/2
Solve 4x² − 12x + 9 = 0.
Use: a² − 2ab + b² = (a − b)²
4x² − 12x + 9 = (2x − 3)² = 0
Answer: x = 3/2, 3/2 (equal roots)
Solve √3x² + 10x + 7√3 = 0.
Given: ac = √3 × 7√3 = 21
Use: Split 10x as 7x + 3x
√3x² + 7x + 3x + 7√3 = x(√3x + 7) + √3(√3x + 7) = (√3x + 7)(x + √3)
Answer: x = −7√3 or x = −√3
Solve x² − (√3 + 1)x + √3 = 0.
Use: Two numbers with sum √3 + 1 and product √3
x² − √3x − x + √3 = x(x − √3) − 1(x − √3) = (x − 1)(x − √3)
Answer: x = 1 or x = √3
Solve 2x² + x − 4 = 0 by the quadratic formula.
Given: a = 2, b = 1, c = −4
Use: x = −b ± √D2a
D = 1 + 32 = 33
x = −1 ± √334
Answer: x = (−1 + √33)/4 or (−1 − √33)/4
Find the nature of the roots of (a) x² − 4x + 4 = 0 (b) 3x² + 2x + 1 = 0 (c) 2x² − 7x + 3 = 0.
Use: D = b² − 4ac
(a) D = 16 − 16 = 0
(b) D = 4 − 12 = −8
(c) D = 49 − 24 = 25
Answer: (a) two equal real roots (b) no real roots (c) two distinct real roots
Find k for which x² − 2kx + 16 = 0 has equal roots.
Given: a = 1, b = −2k, c = 16
Use: D = 0
4k² − 64 = 0 ⇒ k² = 16
Answer: k = 4 or k = −4
Find k for which (k + 1)x² − 2(k − 1)x + 1 = 0 has equal roots.
Given: a = k + 1, b = −2(k − 1), c = 1
Use: D = 0
4(k − 1)² − 4(k + 1) = 0 ⇒ k² − 2k + 1 − k − 1 = 0 ⇒ k² − 3k = 0
k = 0 or k = 3 (both keep a = k + 1 ≠ 0)
Answer: k = 0 or k = 3
For what values of p does x² + 4x + p = 0 have real roots?
Use: Real roots need D ≥ 0
D = 16 − 4p ≥ 0 ⇒ p ≤ 4
Answer: p ≤ 4 (equal roots when p = 4)
Solve 3x² − 2√6x + 2 = 0 using the discriminant.
Given: a = 3, b = −2√6, c = 2
D = 24 − 24 = 0, so the roots are equal
x = −b/2a = 2√66 = √63
Answer: x = √6/3 (twice)
The product of two consecutive odd positive integers is 143. Find them.
Given: integers x and x + 2
x(x + 2) = 143 ⇒ x² + 2x − 143 = 0 ⇒ (x + 13)(x − 11) = 0
x = 11 (positive)
Answer: 11 and 13
The perimeter of a rectangle is 34 m and its diagonal is 13 m. Find its sides.
Given: l + b = 17; diagonal 13
Use: Pythagoras: l² + b² = 13²
(17 − b)² + b² = 169 ⇒ 2b² − 34b + 120 = 0 ⇒ b² − 17b + 60 = 0
(b − 5)(b − 12) = 0 ⇒ b = 5 or 12
Answer: 12 m and 5 m
A train travels 360 km at a uniform speed. Had the speed been 5 km/h more, it would have taken 1 hour less. Find the speed.
Given: speed x km/h
Use: time = distance ÷ speed
360x − 360x + 5 = 1 ⇒ 360 × 5 = x(x + 5) ⇒ x² + 5x − 1800 = 0
(x + 45)(x − 40) = 0 ⇒ x = 40
Answer: 40 km/h
A girl is twice as old as her sister. Four years from now, the product of their ages will be 160. Find their present ages.
Given: sister x, girl 2x
(2x + 4)(x + 4) = 160 ⇒ 2x² + 12x + 16 = 160 ⇒ x² + 6x − 72 = 0
(x + 12)(x − 6) = 0 ⇒ x = 6
Answer: Sister 6 years, girl 12 years
Some students planned a picnic with a budget of ₹480. Eight of them dropped out, so the cost for each of the others went up by ₹10. How many students planned the picnic?
Given: x students
Use: share = 480 ÷ number of students
480x − 8 − 480x = 10 ⇒ 480 × 8 = 10x(x − 8) ⇒ x² − 8x − 384 = 0
(x − 24)(x + 16) = 0 ⇒ x = 24
Answer: 24 students
Is it possible to design a rectangle with perimeter 28 m and area 50 m²?
Given: l + b = 14; lb = 50
Use: Check D
x(14 − x) = 50 ⇒ x² − 14x + 50 = 0
D = 196 − 200 = −4 < 0
Answer: No; the equation has no real roots
In a right triangle, the hypotenuse is 1 m less than twice the shortest side, and the third side is 1 m more than the shortest side. Find the sides.
Given: shortest side x
Use: Pythagoras
x² + (x + 1)² = (2x − 1)² ⇒ 2x² + 2x + 1 = 4x² − 4x + 1 ⇒ 2x² − 6x = 0
2x(x − 3) = 0 ⇒ x = 3 (x = 0 is not a side)
Answer: 3 m, 4 m and 5 m
Pick a level and answer 10 questions with instant feedback. Each level has a bank of 20 questions, and New 10 questions draws a fresh set.
Read the Assertion (A) and the Reason (R), then pick one of the four standard options. You get 5 questions per set from a bank of 20 at each level, and New 5 questions draws a fresh set.
Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.
(i) x² + 2x + 1 = 2x − 6 ⇒ x² + 7 = 0. Quadratic.
(ii) x² − 2x = −6 + 2x ⇒ x² − 4x + 6 = 0. Quadratic.
(iii) x² − x − 2 = x² + 2x − 3 ⇒ −3x + 1 = 0. Not quadratic (degree 1).
(iv) 2x² − 5x − 3 = x² + 5x ⇒ x² − 10x − 3 = 0. Quadratic.
(v) 2x² − 7x + 3 = x² + 4x − 5 ⇒ x² − 11x + 8 = 0. Quadratic.
(vi) x² + 3x + 1 = x² − 4x + 4 ⇒ 7x − 3 = 0. Not quadratic.
(vii) x³ + 6x² + 12x + 8 = 2x³ − 2x ⇒ x³ − 6x² − 14x − 8 = 0. Not quadratic (degree 3).
(viii) x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8 ⇒ 2x² − 13x + 9 = 0. Quadratic.
Answer: Quadratic: (i), (ii), (iv), (v), (viii). Not quadratic: (iii), (vi), (vii)
(i) Breadth x m, length (2x + 1) m: x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0.
(ii) Integers x and x + 1: x(x + 1) = 306 ⇒ x² + x − 306 = 0.
(iii) Rohan x years, mother (x + 26) years. In 3 years: (x + 3)(x + 29) = 360 ⇒ x² + 32x + 87 = 360 ⇒ x² + 32x − 273 = 0.
(iv) Speed x km/h. Time = 480/x h; at (x − 8) km/h, time = 480/(x − 8) h. 480x − 8 − 480x = 3 ⇒ 480x − 480(x − 8) = 3x(x − 8) ⇒ 3840 = 3x² − 24x ⇒ x² − 8x − 1280 = 0.
Answer: (i) 2x² + x − 528 = 0 (ii) x² + x − 306 = 0 (iii) x² + 32x − 273 = 0 (iv) x² − 8x − 1280 = 0
(i) x² − 5x + 2x − 10 = x(x − 5) + 2(x − 5) = (x − 5)(x + 2) = 0 ⇒ x = 5, −2.
(ii) 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2) = 0 ⇒ x = 3/2, −2.
(iii) √2x² + 5x + 2x + 5√2 = x(√2x + 5) + √2(√2x + 5) = (√2x + 5)(x + √2) = 0 ⇒ x = −5√2 = −5√22, −√2.
(iv) Multiply by 8: 16x² − 8x + 1 = 0 ⇒ (4x − 1)² = 0 ⇒ x = 1/4, 1/4.
(v) 100x² − 10x − 10x + 1 = 10x(10x − 1) − 1(10x − 1) = (10x − 1)² = 0 ⇒ x = 1/10, 1/10.
Answer: (i) 5, −2 (ii) 3/2, −2 (iii) −5/√2, −√2 (iv) 1/4, 1/4 (v) 1/10, 1/10
(i) x² − 45x + 324 = 0 ⇒ x² − 36x − 9x + 324 = (x − 36)(x − 9) = 0 ⇒ x = 36 or 9.
If John had 36, Jivanti had 9; if John had 9, Jivanti had 36. Check: (36 − 5)(9 − 5) = 31 × 4 = 124 ✓
(ii) x² − 55x + 750 = 0 ⇒ x² − 25x − 30x + 750 = (x − 25)(x − 30) = 0 ⇒ x = 25 or 30.
Check: 25 × (55 − 25) = 750 and 30 × (55 − 30) = 750 ✓
Answer: (i) John 36 and Jivanti 9, or John 9 and Jivanti 36 (ii) 25 toys or 30 toys
Let the numbers be x and 27 − x. x(27 − x) = 182 ⇒ x² − 27x + 182 = 0.
x² − 13x − 14x + 182 = (x − 13)(x − 14) = 0 ⇒ x = 13 or 14. The other number is 14 or 13.
Answer: 13 and 14
Let the integers be x and x + 1. x² + (x + 1)² = 365 ⇒ 2x² + 2x − 364 = 0 ⇒ x² + x − 182 = 0.
(x + 14)(x − 13) = 0 ⇒ x = 13 (positive). Check: 169 + 196 = 365 ✓
Answer: 13 and 14
Let the base be x cm; altitude (x − 7) cm. By Pythagoras: x² + (x − 7)² = 13².
2x² − 14x + 49 = 169 ⇒ x² − 7x − 60 = 0 ⇒ (x − 12)(x + 5) = 0 ⇒ x = 12 (a length is positive).
Altitude = 12 − 7 = 5 cm. Check: 5² + 12² = 169 = 13² ✓
Answer: Base 12 cm, altitude 5 cm
Let x articles be produced; each costs ₹(2x + 3). x(2x + 3) = 90 ⇒ 2x² + 3x − 90 = 0.
2x² + 15x − 12x − 90 = x(2x + 15) − 6(2x + 15) = (2x + 15)(x − 6) = 0 ⇒ x = 6 (cannot be negative).
Cost of each article = 2 × 6 + 3 = ₹15.
Answer: 6 articles, ₹15 each
(i) a = 2, b = −3, c = 5: D = 9 − 40 = −31 < 0. No real roots.
(ii) a = 3, b = −4√3, c = 4: D = 48 − 48 = 0. Two equal real roots, each −b/2a = 4√36 = 2√3.
(iii) a = 2, b = −6, c = 3: D = 36 − 24 = 12 > 0. Two distinct real roots: x = 6 ± √124 = 3 ± √32.
Answer: (i) no real roots (ii) equal roots 2/√3, 2/√3 (iii) distinct roots (3 + √3)/2 and (3 − √3)/2
(i) D = k² − 4 × 2 × 3 = k² − 24 = 0 ⇒ k = ±√24 = ±2√6.
(ii) kx² − 2kx + 6 = 0. D = 4k² − 24k = 4k(k − 6) = 0 ⇒ k = 0 or 6.
k = 0 makes the equation 6 = 0, which is not quadratic, so k = 6.
Answer: (i) k = ±2√6 (ii) k = 6
Let the breadth be x m; length 2x m. 2x² = 800 ⇒ x² − 400 = 0.
D = 0 + 1600 > 0, so real roots exist: x = ±20. The breadth is 20 m (positive).
Answer: Yes; length 40 m and breadth 20 m
Let one friend be x years old; the other is (20 − x). Four years ago: (x − 4)(16 − x) = 48.
−x² + 20x − 64 = 48 ⇒ x² − 20x + 112 = 0.
D = 400 − 448 = −48 < 0, so there are no real roots.
Answer: Not possible
Length + breadth = 40 m. Let the length be x m; breadth (40 − x) m. x(40 − x) = 400 ⇒ x² − 40x + 400 = 0.
D = 1600 − 1600 = 0, so there are equal real roots: x = 40/2 = 20. Breadth = 40 − 20 = 20 m.
Answer: Yes; it is a square park 20 m by 20 m