EquationsMCQMTP Nov 20Question 1075 of 221
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If one root of a equation is 2+5\displaystyle 2 + \sqrt{5}, then the quadratic equation is

Options

Ax2−4x−1=0\displaystyle x^2 - 4x - 1 = 0
Bx2−4x+1=0\displaystyle x^2 - 4x + 1 = 0
Cx2+4x+1=0\displaystyle x^2 + 4x + 1 = 0
DNone of these
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Correct Answer

✅ Option a — x2−4x−1=0\displaystyle x^2 - 4x - 1 = 0

All Options:

  • Ax2−4x−1=0\displaystyle x^2 - 4x - 1 = 0
  • Bx2−4x+1=0\displaystyle x^2 - 4x + 1 = 0
  • Cx2+4x+1=0\displaystyle x^2 + 4x + 1 = 0
  • DNone of these

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Detailed Solution & Explanation

If one root of a quadratic equation with rational coefficients is irrational, say α=2+5\displaystyle \alpha = 2 + \sqrt{5}, then the other root β\displaystyle \beta must be its irrational conjugate:
β=2−5\beta = 2 - \sqrt{5}
Now, we calculate the sum and product of the roots:
1. **Sum of roots (S\displaystyle S):**
S=α+β=(2+5)+(2−5)=4S = \alpha + \beta = (2 + \sqrt{5}) + (2 - \sqrt{5}) = 4
2. **Product of roots (P\displaystyle P):**
P=αβ=(2+5)(2−5)=(2)2−(5)2=4−5=−1P = \alpha\beta = (2 + \sqrt{5})(2 - \sqrt{5}) = (2)^2 - (\sqrt{5})^2 = 4 - 5 = -1
The general form of a quadratic equation is:
x2−Sx+P=0x^2 - Sx + P = 0
Substitute the sum (S=4\displaystyle S = 4) and product (P=−1\displaystyle P = -1):
x2−4x−1=0x^2 - 4x - 1 = 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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