EquationsMCQMTP Dec 23 - Series IIQuestion 1097 of 221
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If the ratio of the roots of the Equation 4x2−6x+p=0\displaystyle 4x^2 - 6x + p = 0 is 1:2\displaystyle 1:2. Then the value of p\displaystyle p is:

Options

A1\displaystyle 1
B2\displaystyle 2
C−2\displaystyle -2
D−1\displaystyle -1
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Correct Answer

✅ Option b — 2\displaystyle 2

All Options:

  • A1\displaystyle 1
  • B2\displaystyle 2
  • C−2\displaystyle -2
  • D−1\displaystyle -1

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Detailed Solution & Explanation

Let the roots of the quadratic equation 4x2−6x+p=0\displaystyle 4x^2 - 6x + p = 0 be α\displaystyle \alpha and 2α\displaystyle 2\alpha, since the ratio of the roots is given as 1:2\displaystyle 1:2.

For a quadratic equation Ax2+Bx+C=0\displaystyle Ax^2 + Bx + C = 0, the sum of roots is −B/A\displaystyle -B/A and the product of roots is C/A\displaystyle C/A.
Here, A=4\displaystyle A = 4, B=−6\displaystyle B = -6, and C=p\displaystyle C = p.

1. **Sum of the roots:**
α+2α=−(−6)4\alpha + 2\alpha = \frac{-(-6)}{4}
3α=64=323\alpha = \frac{6}{4} = \frac{3}{2}
α=12\alpha = \frac{1}{2}

Thus, the roots are α=12\displaystyle \alpha = \frac{1}{2} and 2α=1\displaystyle 2\alpha = 1.

2. **Product of the roots:**
α⋅2α=p4\alpha \cdot 2\alpha = \frac{p}{4}
2α2=p42\alpha^2 = \frac{p}{4}
Substituting α=12\displaystyle \alpha = \frac{1}{2}:
2(12)2=p42\left(\frac{1}{2}\right)^2 = \frac{p}{4}
2(14)=p42\left(\frac{1}{4}\right) = \frac{p}{4}
12=p4\frac{1}{2} = \frac{p}{4}
p=2p = 2

Thus, the value of p\displaystyle p is 2\displaystyle 2.

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

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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