If alpha and beta are the roots of 9x^2 + 6x | vedclass

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

Question

(C) Given equation is $9x^2 + 6x + 1 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roo

Vedclass · Accepted Answer

$x^2 + 6x + 9 = 0$

Vedclass · Answer

(C) Given equation is $9x^2 + 6x + 1 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -b/a$ and the product of roots $\alpha\beta = c/a$.Here,$\alpha + \beta = -6/9 = -2/3$ and $\alpha\beta = 1/9$.We need to find the equation whose roots are $1/\alpha$ and $1/\beta$.The sum of the new roots is $S = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-2/3}{1/9} = -6$.The product of the new roots is $P = \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{1/9} = 9$.The required quadratic equation is $x^2 - Sx + P = 0$.Substituting the values,we get $x^2 - (-6)x + 9 = 0$,which simplifies to $x^2 + 6x + 9 = 0$.

If $\alpha$ and $\beta$ are the roots of $9x^2 + 6x + 1 = 0$,then the equation with the roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is

Similar Questions

Solve the given two equations and select the correct option.
$I.$ $2x^2 + 23x + 63 = 0$
$II.$ $4y^2 + 19y + 21 = 0$

How many roots does the equation $x - \frac{2}{x - 1} = 1 - \frac{2}{x - 1}$ have?

The quadratic equation in $t$,such that the Arithmetic Mean $(A.M.)$ of its roots is $A$ and the Geometric Mean $(G.M.)$ is $G$,is

If the ratio of the roots of the equation $ax^2 + bx + c = 0$ is $p:q$,then:

The roots of the equation ${x^4} - 8{x^2} - 9 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module