If α and β are the zeroes of polynomial 3x² + 6x + k such that α + β + αβ = -2/3, then the value of k is : (A) -8 (B) 8 (C) -4 (D) 4

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/1/1 · 1 mark · MCQ

CBSE Class 10 Mathematics Standard 2025: question 1

Set in the CBSE Class 10 board exam of 2025, paper 30/1/1, section A, question 1, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q1 · 1 mark · MCQPolynomials

If $α$ and $β$ are the zeroes of polynomial $3 x^{2} + 6 x + k$ such that $α + β + α β = - \frac{2}{3}$ , then the value of k is :

(A) $- 8$
(B)8
(C) $- 4$
(D)4

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Recall the formulas for the sum and the product of the zeroes of $a x^{2} + b x + c$ .

Hint 2

Here $α + β = - 2$ and $α β = k /3$ . Put these into the given condition.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

For $3 x^{2} + 6 x + k$ : $a = 3, b = 6, c = k$ . So $α + β = - \frac{b}{a} = - \frac{6}{3} = - 2$ and $α β = \frac{c}{a} = \frac{k}{3}$ .
Substitute in the given condition: $- 2 + \frac{k}{3} = - \frac{2}{3}$ , so $\frac{k}{3} = 2 - \frac{2}{3} = \frac{4}{3}$ .
Hence $k = 4$ , which is option (D).
1 mark

Answer: k = 4, option (D)

Where the marks are

Uses sum $= - 2$ and product $= k /3$ to get $k = 4$ , option (D) (1 mark)

Revise the chapter

Chapter: Polynomials (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Polynomials, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 30/1/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 1

Hints

Step-by-step solution

Revise the chapter

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