If alpha and beta are the roots of the equati | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$. Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. The roots of t

Vedclass · Accepted Answer

$a+b+c$

Vedclass · Answer

(C) Given that $\alpha$ and $\beta$ are the roots of $ax^2+bx+c=0$. Then,$\alpha+\beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$. The roots of the new equation are $\gamma = \frac{1-\alpha}{\alpha} = \frac{1}{\alpha}-1$ and $\delta = \frac{1-\beta}{\beta} = \frac{1}{\beta}-1$. The new equation is $x^2 - (\gamma+\delta)x + \gamma\delta = 0$. Sum of roots: $\gamma+\delta = (\frac{1}{\alpha}-1) + (\frac{1}{\beta}-1) = \frac{\alpha+\beta}{\alpha\beta} - 2 = \frac{-b/a}{c/a} - 2 = -\frac{b}{c} - 2 = -\frac{b+2c}{c}$. Product of roots: $\gamma\delta = (\frac{1}{\alpha}-1)(\frac{1}{\beta}-1) = \frac{1}{\alpha\beta} - (\frac{1}{\alpha}+\frac{1}{\beta}) + 1 = \frac{1}{\alpha\beta} - \frac{\alpha+\beta}{\alpha\beta} + 1 = \frac{a}{c} - \frac{-b/a}{c/a} + 1 = \frac{a}{c} + \frac{b}{c} + 1 = \frac{a+b+c}{c}$. The equation is $x^2 - (-\frac{b+2c}{c})x + \frac{a+b+c}{c} = 0$. Multiplying by $c$,we get $cx^2 + (b+2c)x + (a+b+c) = 0$. Comparing with $px^2+qx+r=0$,we find $r = a+b+c$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2+bx+c=0$ and the equation having roots $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ is $px^2+qx+r=0$,then $r=$

Similar Questions

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + k,$ where $k$ is a constant of integration,then $A + B + C$ equals

The lengths of tangent,subtangent,normal,and subnormal for the curve $y=x^2+x-1$ at $(1,1)$ are $A, B, C$,and $D$ respectively. Then their increasing order is:

Which of the following $0.05 \, M$ aqueous solution will have the lowest freezing point?

$A$ body falling from a height of $10\,m$ rebounds from a hard floor. If it loses $20\%$ of its energy on impact,then the coefficient of restitution is:

The charge on colloidal particles is due to

Vedclass Test Series

Exam Paper Generator

Online Exam Module