If the roots of the equation ax^2 + bx + c = | vedclass

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

Question

(A) Let the roots be $x_1 = \frac{\alpha}{\alpha - 1}$ and $x_2 = \frac{\alpha + 1}{\alpha}$.From the relation between roots and coefficients,the sum

Vedclass · Accepted Answer

$b^2 - 4ac$

Vedclass · Answer

(A) Let the roots be $x_1 = \frac{\alpha}{\alpha - 1}$ and $x_2 = \frac{\alpha + 1}{\alpha}$.From the relation between roots and coefficients,the sum of roots is $x_1 + x_2 = -\frac{b}{a}$ and the product of roots is $x_1 x_2 = \frac{c}{a}$.Product of roots: $\frac{\alpha}{\alpha - 1} 	imes \frac{\alpha + 1}{\alpha} = \frac{\alpha + 1}{\alpha - 1} = \frac{c}{a}$.By componendo and dividendo,$\frac{(\alpha + 1) + (\alpha - 1)}{(\alpha + 1) - (\alpha - 1)} = \frac{c + a}{c - a} \implies \frac{2\alpha}{2} = \frac{c + a}{c - a} \implies \alpha = \frac{c + a}{c - a}$.Sum of roots: $\frac{\alpha^2 + \alpha^2 - 1}{\alpha(\alpha - 1)} = \frac{2\alpha^2 - 1}{\alpha^2 - \alpha} = -\frac{b}{a}$.Substituting $\alpha$ and simplifying,we find that $(a + b + c)^2 = b^2 - 4ac$.

If the roots of the equation $ax^2 + bx + c = 0$ are real and of the form $\frac{\alpha}{\alpha - 1}$ and $\frac{\alpha + 1}{\alpha}$,then the value of $(a + b + c)^2$ is

Similar Questions

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $2x^3 - 3x^2 + 6x + 1 = 0$,then $\alpha^2 + \beta^2 + \gamma^2$ is equal to

In a triangle $PQR$,$\angle R = \frac{\pi}{2}$. If $\tan(\frac{P}{2})$ and $\tan(\frac{Q}{2})$ are the roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

If $\frac{x-4}{x^2-5x-2k} = \frac{2}{x-2} - \frac{1}{x+k}$,then $k$ is equal to

If $x^2 + px + 1$ is a factor of $ax^3 + bx + c$,then:

If $\alpha$ and $\beta$ are roots of $ax^2 + 2bx + c = 0$,then $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module