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

The sum of the solutions of the equation $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $(x > 0)$ is equal to

Solve the given equations and select the correct option.
$I.$ $8x + 7y = 135$
$II.$ $5x + 6y = 99$
$III.$ $9y + 8z = 121$

Let $\alpha, \beta, \gamma, \delta$ be the roots of the equation $x^4 + x^2 + 1 = 0$. Then,the equation whose roots are $\alpha^2, \beta^2, \gamma^2, \delta^2$ is:

The solution set of the equation $x^{\log_x(1 - x)^2} = 9$ is

When $a = -5, b = -6$ and $c = 10$,find the value of $\frac{a^{3} + b^{3} + c^{3} - 3abc}{ab + bc + ca - a^{2} - b^{2} - c^{2}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module