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

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

Solve the given two equations and select the correct answer from the given options.
$I.$ $\sqrt{784} x + 1234 = 1486$
$II.$ $\sqrt{1089} y + 2081 = 2345$

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is equal to

Solve the given two equations and select the correct answer from the given options.
$I.$ $7x^{2} + 16x - 15 = 0$
$II.$ $y^{2} - 6y - 7 = 0$

If the sum of the roots of the equation ${x^2} + px + q = 0$ is equal to the sum of their squares,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module