EquationsMCQPYQ Nov 18Question 1053 of 221
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When two roots of QE are a,1a\displaystyle a, \frac{1}{a} then what will be the quadratic equation?

Options

Aax2−(a2+1)x+a=0\displaystyle ax^2 - (a^2+1)x + a = 0
Bax2−a2x+1=0\displaystyle ax^2 - a^2x + 1 = 0
Cax2−(a2+1)x−1=0\displaystyle ax^2 - (a^2+1)x - 1 = 0
DNone of these
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Correct Answer

✅ Option a — ax2−(a2+1)x+a=0\displaystyle ax^2 - (a^2+1)x + a = 0

All Options:

  • Aax2−(a2+1)x+a=0\displaystyle ax^2 - (a^2+1)x + a = 0
  • Bax2−a2x+1=0\displaystyle ax^2 - a^2x + 1 = 0
  • Cax2−(a2+1)x−1=0\displaystyle ax^2 - (a^2+1)x - 1 = 0
  • DNone of these

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Detailed Solution & Explanation

Given the roots of the quadratic equation:
α=a\displaystyle \alpha = a and β=1a\displaystyle \beta = \frac{1}{a}

The standard form of a quadratic equation is:
x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0
Substitute the sum and product of the roots:
α+β=a+1a=a2+1a\alpha + \beta = a + \frac{1}{a} = \frac{a^2 + 1}{a}
αβ=a⋅1a=1\alpha\beta = a \cdot \frac{1}{a} = 1

Now construct the equation:
x2−(a2+1a)x+1=0x^2 - \left(\frac{a^2 + 1}{a}\right)x + 1 = 0
Multiply the entire equation by a\displaystyle a to remove the denominator:
ax2−(a2+1)x+a=0ax^2 - (a^2 + 1)x + a = 0
This matches Option a.
**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.

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