Question:
If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 - 3x + 1 = 0 \), then \( \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta} + \frac{\gamma}{1-\gamma} = \)
If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 - 3x + 1 = 0 \), then \( \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta} + \frac{\gamma}{1-\gamma} = \)
Show Hint
Transformation of equations is the most robust technique for finding sums of rational functions of roots. It avoids complex algebra and symmetric sum calculations.
Updated On: Jun 9, 2026
- \( 0 \)
- \( 3 \)
- \( -3 \)
- \( 1 \)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
While one could calculate the symmetric sum using Vieta's formulas, the method of transformation of equations is significantly more efficient. We aim to find the sum of roots of a new cubic equation where the roots are of the form \( y = \frac{x}{1-x} \).
Step 1: Establish the transformation relationship.
Let \( y = \frac{x}{1-x} \). We need to express \( x \) in terms of \( y \): \( y(1-x) = x \Rightarrow y - yx = x \Rightarrow y = x(1+y) \Rightarrow x = \frac{y}{1+y} \).
Step 2: Substitute into the original equation.
Given \( x^3 - 3x + 1 = 0 \), substitute \( x = \frac{y}{1+y} \): \[ \left( \frac{y}{1+y} \right)^3 - 3\left( \frac{y}{1+y} \right) + 1 = 0 \]
Step 3: Simplify the equation to standard polynomial form.
Multiply by \( (1+y)^3 \): \[ y^3 - 3y(1+y)^2 + (1+y)^3 = 0 \] Expand the terms: \[ y^3 - 3y(1 + 2y + y^2) + (1 + 3y + 3y^2 + y^3) = 0 \] \[ y^3 - 3y - 6y^2 - 3y^3 + 1 + 3y + 3y^2 + y^3 = 0 \] \[ (y^3 - 3y^3 + y^3) + (-6y^2 + 3y^2) + (-3y + 3y) + 1 = 0 \Rightarrow -y^3 - 3y^2 + 1 = 0 \] This simplifies to: \( y^3 + 3y^2 - 1 = 0 \).
Step 4: Find the sum of roots.
For \( y^3 + 3y^2 + 0y - 1 = 0 \), the sum of roots \(\sum \frac{\alpha}{1-\alpha}\) is \( - (\text{coefficient of } y^2) / (\text{coefficient of } y^3) = -3/1 = -3 \). -3
Step 1: Establish the transformation relationship.
Let \( y = \frac{x}{1-x} \). We need to express \( x \) in terms of \( y \): \( y(1-x) = x \Rightarrow y - yx = x \Rightarrow y = x(1+y) \Rightarrow x = \frac{y}{1+y} \).
Step 2: Substitute into the original equation.
Given \( x^3 - 3x + 1 = 0 \), substitute \( x = \frac{y}{1+y} \): \[ \left( \frac{y}{1+y} \right)^3 - 3\left( \frac{y}{1+y} \right) + 1 = 0 \]
Step 3: Simplify the equation to standard polynomial form.
Multiply by \( (1+y)^3 \): \[ y^3 - 3y(1+y)^2 + (1+y)^3 = 0 \] Expand the terms: \[ y^3 - 3y(1 + 2y + y^2) + (1 + 3y + 3y^2 + y^3) = 0 \] \[ y^3 - 3y - 6y^2 - 3y^3 + 1 + 3y + 3y^2 + y^3 = 0 \] \[ (y^3 - 3y^3 + y^3) + (-6y^2 + 3y^2) + (-3y + 3y) + 1 = 0 \Rightarrow -y^3 - 3y^2 + 1 = 0 \] This simplifies to: \( y^3 + 3y^2 - 1 = 0 \).
Step 4: Find the sum of roots.
For \( y^3 + 3y^2 + 0y - 1 = 0 \), the sum of roots \(\sum \frac{\alpha}{1-\alpha}\) is \( - (\text{coefficient of } y^2) / (\text{coefficient of } y^3) = -3/1 = -3 \). -3
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET Polynomials Questions
- The equation \( x^4 - x^3 - 6x^2 + 4x + 8 = 0 \) has two equal roots. If \( \alpha, \beta \) are the other two roots of this equation, then \( \alpha^2 + \beta^2 = \):
- The condition that the roots of \( x^3 - bx^2 + cx - d = 0 \) are in arithmetic progression is:
If the sum of two roots of \( x^3 + px^2 + qx - 5 = 0 \) is equal to its third root, then \( p(q^2 - 4q) = \) ?
- If \( P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e \) is a polynomial such that: \[ P(0) = 1, \quad P(1) = 2, \quad P(2) = 5, \] \[ P(3) = 10, \quad P(4) = 17, \] then find the value of \( P(5) \)=
- If $ 3 + i $ and $ 2 - \sqrt{3} $ are the roots of the equation $ f(x) = a_0 + a_1 x + a_2 x^2 + ... + a_n x^n $, where $ a_0, a_1, ..., a_n \in \mathbb{Z} $, then the least value of $ n $ and the value of $ a_0 $ are respectively:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions