15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
Which of the following is not a quadratic equation?
Why: (x − 3)² = x² + 1 simplifies to −6x + 8 = 0, which is linear.
\((x-3)^2 = x^2 + 1 \Rightarrow x^2 - 6x + 9 = x^2 + 1 \Rightarrow -6x + 8 = 0\): degree \(1\). The others reduce to degree \(2\).
The roots of \(x^2 - 11x + 28 = 0\) are
Why: 28 = 4 × 7 and 4 + 7 = 11: (x − 4)(x − 7) = 0.
\(x^2 - 11x + 28 = (x - 4)(x - 7) = 0 \Rightarrow x = 4\) or \(x = 7\).
The discriminant of \(2x^2 - 5x + 4 = 0\) is
Why: D = b² − 4ac = 25 − 32.
\(D = (-5)^2 - 4(2)(4) = 25 - 32 = -7\).
The equation \(x^2 + 5x + 7 = 0\) has
Why: D = 25 − 28 = −3 < 0.
\(D = 5^2 - 4(1)(7) = -3 \lt 0\), so there are no real roots.
Using the quadratic formula, the roots of \(x^2 - 6x + 4 = 0\) are
Why: x = (6 ± √(36 − 16))/2 = (6 ± 2√5)/2.
\(x = \dfrac{6 \pm \sqrt{36 - 16}}{2} = \dfrac{6 \pm 2\sqrt5}{2} = 3 \pm \sqrt5\).
The roots of \(x^2 + px + q = 0\) are \(3\) and \(-5\). Then \(p + q\) equals
Why: The equation is (x − 3)(x + 5) = x² + 2x − 15 = 0, so p = 2, q = −15.
A quadratic with roots \(3\) and \(-5\) is \((x - 3)(x + 5) = x^2 + 2x - 15\). So \(p = 2\), \(q = -15\) and \(p + q = -13\).
The roots of \(2x^2 - 3x - 5 = 0\) are
Why: Split −3x as −5x + 2x: (2x − 5)(x + 1) = 0.
\(2x^2 - 5x + 2x - 5 = x(2x - 5) + 1(2x - 5) = (2x - 5)(x + 1) = 0\), so \(x = \dfrac52\) or \(x = -1\).
The equation \(2x^2 + 4x + k = 0\) has no real roots exactly when
Why: No real roots needs D = 16 − 8k < 0.
\(D = 16 - 8k \lt 0 \Rightarrow k \gt 2\). At \(k = 2\), \(D = 0\) (equal roots).
The product of two consecutive positive integers is \(132\). The integers are
Why: n(n + 1) = 132 gives n² + n − 132 = 0 = (n + 12)(n − 11).
\(n(n+1) = 132 \Rightarrow n^2 + n - 132 = 0 \Rightarrow (n + 12)(n - 11) = 0\). As \(n \gt 0\), \(n = 11\): the integers are \(11\) and \(12\).
The length of a rectangular plot is \(3\) m more than its breadth and its area is \(70\ \text{m}^2\). The breadth is
Why: b(b + 3) = 70 gives b² + 3b − 70 = 0 = (b + 10)(b − 7).
\(b(b + 3) = 70 \Rightarrow b^2 + 3b - 70 = 0 \Rightarrow (b + 10)(b - 7) = 0\). Breadth is positive, so \(b = 7\) m.
The roots of \((x - 3)(x + 2) = 14\) are
Why: Expand first: x² − x − 20 = 0 = (x − 5)(x + 4).
\(x^2 - x - 6 = 14 \Rightarrow x^2 - x - 20 = 0 \Rightarrow (x - 5)(x + 4) = 0\), so \(x = 5\) or \(-4\). (Setting each bracket to \(0\) is wrong: the right side is not \(0\).)
The values of \(k\) for which \(x^2 - kx + 25 = 0\) has equal roots are
Why: Equal roots need D = k² − 100 = 0.
\(D = (-k)^2 - 4(1)(25) = k^2 - 100 = 0 \Rightarrow k = \pm 10\).
Which of the following equations has two equal roots?
Why: x² − 10x + 25 = (x − 5)², so D = 100 − 100 = 0.
Discriminants: \(16\), \(-3\), \(0\), \(8\). Only \(x^2 - 10x + 25 = 0\) has \(D = 0\) (root \(5\) twice).
The sum of a positive number and its reciprocal is \(\dfrac{17}{4}\). The number is
Why: x + 1/x = 17/4 gives 4x² − 17x + 4 = 0 = (4x − 1)(x − 4).
\(4x^2 + 4 = 17x \Rightarrow 4x^2 - 17x + 4 = 0 \Rightarrow (4x - 1)(x - 4) = 0\), so \(x = 4\) or \(\dfrac14\) (both work).
The product of Tara's age \(5\) years ago and her age \(5\) years from now is \(144\). Her present age is
Why: (x − 5)(x + 5) = 144 gives x² = 169.
\((x - 5)(x + 5) = 144 \Rightarrow x^2 - 25 = 144 \Rightarrow x^2 = 169 \Rightarrow x = 13\) (age is positive).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
Aarav cycles \(36\) km from his town to his grandmother's village at a steady speed of \(x\) km/h. He notices that if he rode \(3\) km/h faster, the trip would take \(1\) hour less.
A shopkeeper spends ₹\(1200\) on \(x\) identical school bags. He notices that if each bag had cost ₹\(20\) less, he could have bought \(5\) more bags for the same ₹\(1200\).
This free set is separate from the chapter's question bank. On the Quadratic Equations chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 39 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.