If alpha and beta are roots of the equation ax2 + bx + c = 0 then the equation whose roots are 1 and 1/beta is:

Option b: cx2 + bx + a = 0. Solution: Note: The question contains a slight typographical error. The roots of the new equation should be 1/alpha and 1/beta (instead of 1 and 1/beta ). For the original equation ax2 + bx + c = 0 : Sum of roots: alpha + beta = -b/a Product of roots: alphabeta = c/a Let the roots of the new equation be y1 = 1/alpha and y2 = 1/beta . The sum of the new roots ( S ) is: S = y1 + y2 = 1/alpha + 1/beta = alpha + beta/alphabeta = -b/a/c/a = -b/c The product of the new roots ( P ) is: P = y1 × y2 = 1/alpha × 1/beta = 1/alphabeta = 1/c/a = a/c The quadratic equation is given by: x2 - Sx + P = 0 Substitute S and P : x2 - (-b/c)x + a/c = 0 x2 + b/cx + a/c = 0 Multiply the entire equation by c : cx2 + bx + a = 0 This corresponds to Option b (and Option a, as they are identical). Option b

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

The correct answer is Option b: $cx^2 + bx + a = 0$. *Note:* The question contains a slight typographical error. The roots of the new equation should be $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ (instead of $1$ and $\frac{1}{\beta}$). For the original equation $ax^2 + bx + c = 0$: Sum of roots: $\alpha + \beta = -\frac{b}{a}$ Product of roots: $\alpha\beta = \frac{c}{a}$ Let the roots of the new equation be $y_1 = \frac{1}{\alpha}$ and $y_2 = \frac{1}{\beta}$. The sum of the new roots ($S$) is: $$S = y_1 + y_2 = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}$$ The product of the new roots ($P$) is: $$P = y_1 \cdot y_2 = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}$$ The quadratic equation is given by: $$x^2 - Sx + P = 0$$ Substitute $S$ and $P$: $$x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} = 0$$ $$x^2 + \frac{b}{c}x + \frac{a}{c} = 0$$ Multiply the entire equation by $c$: $$cx^2 + bx + a = 0$$ This corresponds to Option b (and Option a, as they are identical). **Option b**

EquationsMCQPYQ Jun 24Question 1064 of 221

All Questions

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

Options

\displaystyle cx^2 + bx + a = 0

\displaystyle cx^2 + bx + a = 0

\displaystyle cx^2 - bx + a = 0

\displaystyle x^2 + bx - a = 0

For any discrepancies in this question, email contact@cadada.in

Correct Answer

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

All Options:

A $\displaystyle cx^2 + bx + a = 0$
B $\displaystyle cx^2 + bx + a = 0$
C $\displaystyle cx^2 - bx + a = 0$
D $\displaystyle x^2 + bx - a = 0$

Detailed Solution & Explanation

*Note:* The question contains a slight typographical error. The roots of the new equation should be

\displaystyle \frac{1}{\alpha}

and

\displaystyle \frac{1}{\beta}

(instead of

\displaystyle 1

and

\displaystyle \frac{1}{\beta}

).

For the original equation

\displaystyle ax^2 + bx + c = 0

:
Sum of roots:

\displaystyle \alpha + \beta = -\frac{b}{a}

Product of roots:

\displaystyle \alpha\beta = \frac{c}{a}

Let the roots of the new equation be

\displaystyle y_1 = \frac{1}{\alpha}

and

\displaystyle y_2 = \frac{1}{\beta}

.
The sum of the new roots (

\displaystyle S

) is:

S = y_1 + y_2 = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac{b}{c}

The product of the new roots (

\displaystyle P

) is:

P = y_1 \cdot y_2 = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{c/a} = \frac{a}{c}

The quadratic equation is given by:

x^2 - Sx + P = 0

Substitute

\displaystyle S

and

\displaystyle P

x^2 - \left(-\frac{b}{c}\right)x + \frac{a}{c} = 0

x^2 + \frac{b}{c}x + \frac{a}{c} = 0

Multiply the entire equation by

\displaystyle c

cx^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.

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If α\displaystyle \alphaα and β\displaystyle \betaβ are roots of the equation ax2+bx+c=0\displaystyle ax^2 + bx + c = 0ax2+bx+c=0 then the equation whose roots are 1\displaystyle 11 and 1β\displaystyle \frac{1}{\beta}β1​ is: