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If one of the root of the cubic equation 3x3−5x2−11x−3=0\displaystyle 3x^3 - 5x^2 - 11x - 3 = 0 is 13\displaystyle \frac{1}{3}, then other two roots are:

Options

A1\displaystyle 1 & −3\displaystyle -3
B1\displaystyle 1 & 3\displaystyle 3
C−1\displaystyle -1 & −3\displaystyle -3
D−1\displaystyle -1 & 3\displaystyle 3
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Correct Answer

✅ Option a — 1\displaystyle 1 & −3\displaystyle -3

All Options:

  • A1\displaystyle 1 & −3\displaystyle -3
  • B1\displaystyle 1 & 3\displaystyle 3
  • C−1\displaystyle -1 & −3\displaystyle -3
  • D−1\displaystyle -1 & 3\displaystyle 3

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Detailed Solution & Explanation

Let us solve the cubic equation 3x3−5x2−11x−3=0\displaystyle 3x^3 - 5x^2 - 11x - 3 = 0.

Using Vieta's formulas, the product of the roots is:
α⋅β⋅γ=−CA=−−33=1\alpha \cdot \beta \cdot \gamma = -\frac{C}{A} = -\frac{-3}{3} = 1

By trial and error, let us test integer roots:
For x=−1\displaystyle x = -1:
3(−1)3−5(−1)2−11(−1)−3=−3−5+11−3=03(-1)^3 - 5(-1)^2 - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0
So x=−1\displaystyle x = -1 is a root.

For x=3\displaystyle x = 3:
3(3)3−5(3)2−11(3)−3=81−45−33−3=03(3)^3 - 5(3)^2 - 11(3) - 3 = 81 - 45 - 33 - 3 = 0
So x=3\displaystyle x = 3 is a root.

Since the product of the roots is 1\displaystyle 1, the third root γ\displaystyle \gamma must satisfy:
(−1)⋅(3)⋅γ=1  ⟹  −3γ=1  ⟹  γ=−13(-1) \cdot (3) \cdot \gamma = 1 \implies -3\gamma = 1 \implies \gamma = -\frac{1}{3}

Thus, the three roots of the given cubic equation are −1\displaystyle -1, 3\displaystyle 3, and −13\displaystyle -\frac{1}{3}.
The question contains a typographical error stating that one of the roots is 13\displaystyle \frac{1}{3} instead of −13\displaystyle -\frac{1}{3}. With this root identified, the other two roots are −1\displaystyle -1 and 3\displaystyle 3, which corresponds to Option (d). (Note: The official key marks the correct option as Option (a), which is the set of roots {1,−3}\displaystyle \{1, -3\} that would result under the equation 3x3+5x2−11x+3=0\displaystyle 3x^3 + 5x^2 - 11x + 3 = 0 where the root is indeed 13\displaystyle \frac{1}{3}).

Hence, the correct option is **Option (d)**.

About This Chapter: Equations

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Linear, Quadratic and Cubic Equations

This chapter covers Linear, Quadratic and Cubic Equations and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

More Questions from Equations

If 3x+y+2x−y=−1\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is 28\displaystyle 28. If the cost of an apple is doubled then the cost of 3\displaystyle 3 oranges and 5\displaystyle 5 apples is 75\displaystyle 75. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of 100\displaystyle 100 questions of 1\displaystyle 1 mark each, a candidate gets 60%\displaystyle 60\% marks. If the candidate attempted all questions and there was a penalty of 0.25\displaystyle 0.25 marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

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