EquationsPYQ Sept 25Question 4404 of 155
All Questions

If α\displaystyle \alpha and β\displaystyle \beta are the roots of the equation x2x6=0\displaystyle x^2-x-6=0, then the value of α3+β3+α2+β2+α+β\displaystyle \alpha^3+\beta^3+\alpha^2+\beta^2+\alpha+\beta is equal to

Options

A35
B29
C31
D33
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option d33

All Options:

  • A35
  • B29
  • C31
  • D33

Detailed Solution & Explanation

Given the quadratic equation x2x6=0\displaystyle x^2 - x - 6 = 0 with roots α\displaystyle \alpha and β\displaystyle \beta.
By the relations between roots and coefficients of a quadratic equation: α+β=11=1\alpha + \beta = -\frac{-1}{1} = 1 αβ=61=6\alpha\beta = \frac{-6}{1} = -6
We need to evaluate the expression: E=α3+β3+α2+β2+α+βE = \alpha^3 + \beta^3 + \alpha^2 + \beta^2 + \alpha + \beta
Let us calculate each part of the expression: 1) For α2+β2\displaystyle \alpha^2 + \beta^2: α2+β2=(α+β)22αβ=(1)22(6)=1+12=13\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (1)^2 - 2(-6) = 1 + 12 = 13
2) For α3+β3\displaystyle \alpha^3 + \beta^3: α3+β3=(α+β)(α2αβ+β2)=(1)(13(6))=1×(13+6)=19\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) = (1)(13 - (-6)) = 1 \times (13 + 6) = 19
3) Substitute these values back into the expression: E=19+13+1=33E = 19 + 13 + 1 = 33
Hence, **Option D** is the correct answer.

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

Ready to Master Equations?

Practice all 155 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free