EquationsMCQMTP Nov 18Question 1113 of 221
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If α,β\displaystyle α\displaystyle \alpha, β\displaystyle \beta are the roots of equation x2−4x+5=0\displaystyle x^2 - 4x + 5 = 0 then the equation having roots 1α\displaystyle 1α\displaystyle \frac{1}{\alpha} & 1β\displaystyle 1β\displaystyle \frac{1}{\beta} is

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — $5x^2 - 4x + 1 = 0$

All Options:

  • Ax2−4x+5=0\displaystyle x^2 - 4x + 5 = 0
  • B$4x^2 - 6x + 1 = 0$
  • C$5x^2 - 4x + 1 = 0$
  • D$5x^2 + 4x - 1 = 0$

Detailed Solution & Explanation

Let α\displaystyle α\displaystyle \alpha and β\displaystyle β\displaystyle \beta be the roots of the quadratic equation x2−4x+5=0\displaystyle x^2 - 4x + 5 = 0.
We want to find the quadratic equation whose roots are 1α\displaystyle 1α\displaystyle \frac{1}{\alpha} and 1β\displaystyle 1β\displaystyle \frac{1}{\beta}.

If α\displaystyle α\displaystyle \alpha and β\displaystyle β\displaystyle \beta are the roots of a polynomial equation f(x)=0\displaystyle f(x) = 0, then the equation whose roots are their reciprocals is given by f(1x)=0\displaystyle f\left(1x\displaystyle \frac{1}{x}\right) = 0.

Given:
x2−4x+5=0x^2 - 4x + 5 = 0
Substituting x=1x\displaystyle x = 1x\displaystyle \frac{1}{x}:
(1x)2−4(1x)+5=0\left(1x\displaystyle \frac{1}{x}\right)^2 - 4\left(1x\displaystyle \frac{1}{x}\right) + 5 = 0
1x2−4x+5=01x2−4x+5=0\displaystyle \frac{1}{x^2} - \frac{4}{x} + 5 = 0
Multiplying the entire equation by x2\displaystyle x^2 to clear the fractions:
1−4x+5x2=01 - 4x + 5x^2 = 0
5x2−4x+1=05x^2 - 4x + 1 = 0

Hence, the correct option is **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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