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Quadratic Equations – Class 10 Mathematics notes, important questions and MCQs
NCERT Class 10 Mathematics, chapter 4. Notes, key terms, important questions with answers and MCQs.
Chapter overview
A quadratic equation has the form ax² + bx + c = 0 (a ≠ 0). It is solved by factorisation or the quadratic formula, and the discriminant tells the nature of the roots.
Solving
- Factorisation: split the middle term so that the product equals ac.
- Quadratic formula: x = [−b ± √(b² − 4ac)] / 2a.
- Roots are values of x that satisfy the equation.
Nature of roots (D = b² − 4ac)
- D > 0: two distinct real roots.
- D = 0: two equal real roots, x = −b/2a.
- D < 0: no real roots.
Word problems
- Form the equation from the situation (area, speed, age, numbers).
- Reject roots that do not make sense (negative length, negative speed).
Key terms
- Quadratic equation
- an equation of degree 2, ax² + bx + c = 0, a ≠ 0
- Discriminant
- D = b² − 4ac, decides the nature of the roots
- Root
- a value of x that makes the equation true
Formulas
- x = [−b ± √(b² − 4ac)] / 2a
- D = b² − 4ac
Important questions with answers
Find the nature of roots of 2x² − 4x + 3 = 0.2 marks
D = b² − 4ac = 16 − 24 = −8 < 0, so the equation has no real roots.
Solve 2x² − 7x + 3 = 0 by factorisation.3 marks
2x² − 6x − x + 3 = 2x(x − 3) − 1(x − 3) = (2x − 1)(x − 3) = 0, so x = 1/2 or x = 3.
The product of two consecutive positive integers is 306. Form the quadratic equation and find the integers.5 marks
Let the integers be x and x + 1. Then x(x + 1) = 306, so x² + x − 306 = 0. D = 1 + 1224 = 1225 = 35². x = (−1 ± 35)/2, so x = 17 or −18. Taking the positive value, the integers are 17 and 18. Check: 17 × 18 = 306 ✓.
Quick revision
- ax² + bx + c = 0 with a ≠ 0.
- D > 0 two roots, D = 0 equal roots, D < 0 no real roots.
- Check whether both roots fit a word problem.
Exam tips
- Bring every term to one side (= 0) before factorising.
- For "equal roots" questions put D = 0 and solve for k.
Practice MCQs
Which is a quadratic equation?
- 2x² − 3x + 1 = 0
- x³ − x = 0
- (x + 2)² = x² + 4
- x + 1/x² = 3
Show answer
Answer: (a) 2x² − 3x + 1 = 0. (x + 2)² = x² + 4 reduces to 4x = 0 (linear); only 2x² − 3x + 1 = 0 has degree 2.
Roots of x² − 9 = 0 are
- ±9
- 3 only
- 9
- ±3
Show answer
Answer: (d) ±3. x² = 9 ⇒ x = ±3.
The discriminant of 2x² − 4x + 3 = 0 is
- 8
- 40
- −8
- 16
Show answer
Answer: (c) −8. 16 − 24 = −8.
Roots of x² − 5x + 6 = 0 are
- 2, 3
- −2, −3
- 1, 6
- −1, 6
Show answer
Answer: (a) 2, 3. (x − 2)(x − 3) = 0.
If 2x² + kx + 3 = 0 has equal roots, k =
- 24
- ±6
- ±2√6
- ±√6
Show answer
Answer: (c) ±2√6. k² − 24 = 0 ⇒ k = ±2√6.
x² + 4x + 5 = 0 has
- two distinct real roots
- no real roots
- two equal roots
- one root
Show answer
Answer: (b) no real roots. D = 16 − 20 < 0.
The product of two consecutive positive integers is 306. The smaller integer is
- 16
- 17
- 18
- 15
Show answer
Answer: (b) 17. x(x + 1) = 306 ⇒ x = 17.
A rectangle's length is 1 m more than twice its breadth and its area is 528 m². The breadth is
- 22 m
- 16 m
- 33 m
- 12 m
Show answer
Answer: (b) 16 m. x(2x + 1) = 528 ⇒ 2x² + x − 528 = 0 ⇒ x = 16.
All chapters of Class 10 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.