EquationsMCQPYQ June 19Question 1054 of 221
All Questions

Find the condition that one roots is double the other of ax2+bx+c=0\displaystyle ax^2 + bx + c = 0

Options

A2b2=3ac\displaystyle 2b^2 = 3ac
Bb2=3ac\displaystyle b^2 = 3ac
C2b2=9ac\displaystyle 2b^2 = 9ac
DNone of these
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Correct Answer

✅ Option c — 2b2=9ac\displaystyle 2b^2 = 9ac

All Options:

  • A2b2=3ac\displaystyle 2b^2 = 3ac
  • Bb2=3ac\displaystyle b^2 = 3ac
  • C2b2=9ac\displaystyle 2b^2 = 9ac
  • DNone of these

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Detailed Solution & Explanation

Let the two roots of the quadratic equation ax2+bx+c=0\displaystyle ax^2 + bx + c = 0 be α\displaystyle \alpha and 2α\displaystyle 2\alpha (since one root is double the other).

According to Vieta's formulas:
1) Sum of roots:
α+2α=−ba  ⟹  3α=−ba  ⟹  α=−b3a…(1)\alpha + 2\alpha = -\frac{b}{a} \implies 3\alpha = -\frac{b}{a} \implies \alpha = -\frac{b}{3a} \quad \dots (1)
2) Product of roots:
α⋅2α=ca  ⟹  2α2=ca…(2)\alpha \cdot 2\alpha = \frac{c}{a} \implies 2\alpha^2 = \frac{c}{a} \quad \dots (2)

Substitute the value of α\displaystyle \alpha from equation (1) into equation (2):
2(−b3a)2=ca2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a}
2(b29a2)=ca2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}
2b29a2=ca\frac{2b^2}{9a^2} = \frac{c}{a}
Multiply both sides by 9a2\displaystyle 9a^2:
2b2=9ac2b^2 = 9ac
This is the required condition, which corresponds to Option c.
**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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