EquationsMCQPYQ Jun 24Question 1064 of 221
All Questions

If α\displaystyle α\displaystyle \alpha and β\displaystyle β\displaystyle \beta are roots of the equation ax2+bx+c=0\displaystyle ax^2 + bx + c = 0 then the equation whose roots are 1\displaystyle 1 and 1β\displaystyle 1β\displaystyle \frac{1}{\beta} is:

Select Your Answer (Keyboard: A, B, C, D)
For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — cx2+bx+a=0\displaystyle cx^2 + bx + a = 0

All Options:

  • Acx2+bx+a=0\displaystyle cx^2 + bx + a = 0
  • Bcx2+bx+a=0\displaystyle cx^2 + bx + a = 0
  • Ccx2−bx+a=0\displaystyle cx^2 - bx + a = 0
  • Dx2+bx−a=0\displaystyle x^2 + bx - a = 0

Detailed Solution & Explanation

*Note:* The question contains a slight typographical error. The roots of the new equation should be 1α\displaystyle 1α\displaystyle \frac{1}{\alpha} and 1β\displaystyle 1β\displaystyle \frac{1}{\beta} (instead of 1\displaystyle 1 and 1β\displaystyle 1β\displaystyle \frac{1}{\beta}).

For the original equation ax2+bx+c=0\displaystyle ax^2 + bx + c = 0:
Sum of roots: α+β=−ba\displaystyle α+β\displaystyle \alpha + \beta = -ba\displaystyle \frac{b}{a}
Product of roots: αβ=ca\displaystyle α\displaystyle \alphaβ=ca\displaystyle \beta = \frac{c}{a}

Let the roots of the new equation be y1=1α\displaystyle y_1 = 1α\displaystyle \frac{1}{\alpha} and y2=1β\displaystyle y_2 = 1β\displaystyle \frac{1}{\beta}.
The sum of the new roots (S\displaystyle S) is:
S=y1+y2=1α+1β=α+βαβ=−b/ac/a=−bcS = y_1 + y_2 = \displaystyle \frac{1}{\alpha} + \frac{1}{β\displaystyle \beta} = α+βαβ=−b/ac/a\displaystyle \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -bc\displaystyle \frac{b}{c}
The product of the new roots (P\displaystyle P) is:
P=y1⋅y2=1α⋅1β=1αβ=1c/a=acP = y_1 ⋅\displaystyle \cdot y_2 = 1α\displaystyle \frac{1}{\alpha} ⋅\displaystyle \cdot \displaystyle \frac{1}{\beta} = \frac{1}{α\displaystyle \alphaβ\displaystyle \beta} = 1c/a=ac\displaystyle \frac{1}{c/a} = \frac{a}{c}

The quadratic equation is given by:
x2−Sx+P=0x^2 - Sx + P = 0
Substitute S\displaystyle S and P\displaystyle P:
x2−(−bc)x+ac=0x^2 - \left(-bc\displaystyle \frac{b}{c}\right)x + ac=0\displaystyle \frac{a}{c} = 0
x2+bcx+ac=0x^2 + bc\displaystyle \frac{b}{c}x + ac=0\displaystyle \frac{a}{c} = 0
Multiply the entire equation by c\displaystyle c:
cx2+bx+a=0cx^2 + bx + a = 0
This corresponds to Option b (and Option a, as they are identical).
**Option b**

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

If 3x+y+2x−y=−1\displaystyle 3x+y+2x−y\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = -1 and 1x+y−1x−y=43\displaystyle 1x+y−1x−y=43\displaystyle \frac{1}{x+y} - \frac{1}{x-y} = \frac{4}{3} then (x,y)\displaystyle (x, y) is:

2x+510+3x+1015=5\displaystyle \frac{2x+5}{10} + \frac{3x+10}{15} = 5, find x\displaystyle x

Find value of x2−10x+1\displaystyle x^2 - 10x + 1 if x=15−26\displaystyle x = 15−26\displaystyle \frac{1}{5-2\sqrt{6}}

The cost of 2\displaystyle 2 oranges and 3\displaystyle 3 apples is $28$. If the cost of an apple is doubled then the cost of $3orangesand\displaystyle oranges and5$ apples is $75$. The original cost of 7\displaystyle 7 oranges and 4\displaystyle 4 apples (in Rs) is:

In a multiple choice question paper consisting of $100questionsof\displaystyle questions of1\displaystyle mark each, a candidate gets $60\% marks. If the candidate attempted all questions and there was a penalty of $0.25$ marks for wrong answers is:

The values of x\displaystyle x and y\displaystyle y satisfying the equations 3x+y+2x−y=3\displaystyle 3x+y+2x−y=3\displaystyle \frac{3}{x+y} + \frac{2}{x-y} = 3 and 2x+y+3x−y=233\displaystyle 2x+y+3x−y=233\displaystyle \frac{2}{x+y} + \frac{3}{x-y} = \frac{23}{3} given by

Ready to Master Equations?

Practice all 221 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free