EquationsMCQPYQ Nov 20Question 1056 of 221
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The rational root of the equation 0=2p3−p2−4p+2\displaystyle 0 = 2p^3 - p^2 - 4p + 2 is:

Options

A2
B-2
C12\displaystyle \frac{1}{2}
D14\displaystyle \frac{1}{4}
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Correct Answer

✅ Option c — 12\displaystyle \frac{1}{2}

All Options:

  • A2
  • B-2
  • C12\displaystyle \frac{1}{2}
  • D14\displaystyle \frac{1}{4}

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Detailed Solution & Explanation

Given the cubic equation:
2p3−p2−4p+2=02p^3 - p^2 - 4p + 2 = 0
We can solve this by grouping the terms:
(2p3−p2)−(4p−2)=0(2p^3 - p^2) - (4p - 2) = 0
Factor out the common terms:
p2(2p−1)−2(2p−1)=0p^2(2p - 1) - 2(2p - 1) = 0
(p2−2)(2p−1)=0(p^2 - 2)(2p - 1) = 0
Set each factor to zero to find the roots:
1) p2−2=0  ⟹  p2=2  ⟹  p=±2\displaystyle p^2 - 2 = 0 \implies p^2 = 2 \implies p = \pm\sqrt{2} (these are irrational roots)
2) 2p−1=0  ⟹  2p=1  ⟹  p=12\displaystyle 2p - 1 = 0 \implies 2p = 1 \implies p = \frac{1}{2} (this is a rational root)

Therefore, the rational root is p=12\displaystyle p = \frac{1}{2}, which is Option c.
**Option c**

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.

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