If alpha, beta are the roots of the equation | vedclass

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

Question

(A) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \f

Vedclass · Accepted Answer

$acx^2 + (a + c)bx + (a + c)^2 = 0$

Vedclass · Answer

(A) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \frac{1}{\alpha}$.The sum of the new roots is $S_1 + S_2 = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha \beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.The product of the new roots is $S_1 S_2 = (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha \beta + 1 + 1 + \frac{1}{\alpha \beta} = \alpha \beta + 2 + \frac{1}{\alpha \beta} = \frac{c}{a} + 2 + \frac{a}{c} = \frac{c^2 + 2ac + a^2}{ac} = \frac{(a+c)^2}{ac}$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - [-\frac{b(a+c)}{ac}]x + \frac{(a+c)^2}{ac} = 0$.Multiplying by $ac$,we get $acx^2 + b(a+c)x + (a+c)^2 = 0$.

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

Similar Questions

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the value of $\alpha\beta^2 + \alpha^2\beta + \alpha\beta$ will be

If one root of $5x^2 + 13x + k = 0$ is the reciprocal of the other,then $k = $?

If the roots of the equation $(x - a)(x - b) = c$ (where $c \neq 0$) are $\alpha$ and $\beta$,then what are the roots of the equation $(x - \alpha)(x - \beta) + c = 0$?

If the sum of the roots of the quadratic equation $ax^2 + bx + c = 0$ is equal to the sum of the squares of their reciprocals,then $a/c, b/a, c/b$ are in

The coefficient of $x$ in the equation $x^2 + px + q = 0$ was taken as $17$ in place of $13$. Its roots were found to be $-2$ and $-15$. The roots of the original equation are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module