EquationsMCQMTP Sep 24 Series IIQuestion 1103 of 221
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If α,β\displaystyle \alpha, \beta are the roots of the QE 3x2−4x+1=0\displaystyle 3x^2 - 4x + 1 = 0, the eq. having roots α2β\displaystyle \frac{\alpha^2}{\beta} and β2α\displaystyle \frac{\beta^2}{\alpha} is:

Options

A9x2−28x+3=0\displaystyle 9x^2 - 28x + 3 = 0
B9x2−28x+1=0\displaystyle 9x^2 - 28x + 1 = 0
C9x2−28x+5=0\displaystyle 9x^2 - 28x + 5 = 0
DNone of these
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Correct Answer

✅ Option a — 9x2−28x+3=0\displaystyle 9x^2 - 28x + 3 = 0

All Options:

  • A9x2−28x+3=0\displaystyle 9x^2 - 28x + 3 = 0
  • B9x2−28x+1=0\displaystyle 9x^2 - 28x + 1 = 0
  • C9x2−28x+5=0\displaystyle 9x^2 - 28x + 5 = 0
  • DNone of these

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Detailed Solution & Explanation

For the quadratic equation 3x2−4x+1=0\displaystyle 3x^2 - 4x + 1 = 0, let the roots be α\displaystyle \alpha and β\displaystyle \beta.
Using Vieta's formulas:
1. Sum of roots: α+β=43\displaystyle \alpha + \beta = \frac{4}{3}
2. Product of roots: αβ=13\displaystyle \alpha \beta = \frac{1}{3}

We want to find the quadratic equation whose roots are R1=α2β\displaystyle R_1 = \frac{\alpha^2}{\beta} and R2=β2α\displaystyle R_2 = \frac{\beta^2}{\alpha}.

1. **Sum of the new roots (S\displaystyle S):**
S=R1+R2=α2β+β2α=α3+β3αβS = R_1 + R_2 = \frac{\alpha^2}{\beta} + \frac{\beta^2}{\alpha} = \frac{\alpha^3 + \beta^3}{\alpha \beta}
Using the algebraic identity α3+β3=(α+β)3−3αβ(α+β)\displaystyle \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta):
α3+β3=(43)3−3(13)(43)\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)^3 - 3\left(\frac{1}{3}\right)\left(\frac{4}{3}\right)
α3+β3=6427−43=64−3627=2827\alpha^3 + \beta^3 = \frac{64}{27} - \frac{4}{3} = \frac{64 - 36}{27} = \frac{28}{27}
Substitute this back into the sum expression:
S=282713=2827×3=289S = \frac{\frac{28}{27}}{\frac{1}{3}} = \frac{28}{27} \times 3 = \frac{28}{9}

2. **Product of the new roots (P\displaystyle P):**
P=R1⋅R2=(α2β)(β2α)=αβ=13P = R_1 \cdot R_2 = \left(\frac{\alpha^2}{\beta}\right)\left(\frac{\beta^2}{\alpha}\right) = \alpha \beta = \frac{1}{3}

3. **Forming the quadratic equation:**
The required quadratic equation is given by:
x2−Sx+P=0x^2 - Sx + P = 0
x2−289x+13=0x^2 - \frac{28}{9}x + \frac{1}{3} = 0
Multiplying the entire equation by 9\displaystyle 9 to clear the fractions:
9x2−28x+3=09x^2 - 28x + 3 = 0

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

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

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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:

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