For the equation ax^2 + bx + c = 0,if alpha a | 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 + \fr

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 $S_2 = \beta + \frac{1}{\alpha}$.Sum of roots $= (\alpha + \beta) + (\frac{1}{\beta} + \frac{1}{\alpha}) = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.Product of roots $= (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = 2 + \frac{c}{a} + \frac{a}{c} = \frac{2ac + c^2 + a^2}{ac} = \frac{(a+c)^2}{ac}$.The required 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 $acx^2 + (a+c)bx + (a+c)^2 = 0$.

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

Similar Questions

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

If $\tan A$ and $\tan B$ are the roots of the equation $5x^2 - 4x + 1 = 0$, then the value of $A + B$ is...

If one root of the equation $ax^3+bx+c=0$ is twice another root,then

If $-1$ is a twice repeated root of the equation $ax^3+bx^2+cx+1=0$,then

If the roots of the equation $ax^2 + bx + c = 0$ are $\sin \alpha$ and $\cos \alpha$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module