Question

if alpha and beta are the zeroes of 9x^2 -9x+2 find value of 1/alpha and 1/beta .... and please solve this answer in proper way​

Answer 1

Given :-

• 9x² - 9x + 2 has roots α, β are the roots of Quadratic equation

To find:-

Value of 1/β and 1/α

Solution:-

9x² - 9x + 2 First we will find the roots that means α, β We can find roots of The Quadratic equation by factorisation method

$9x {}^{2} - 9x + 2 = 0$

Here sum of roots should be -9x and product of roots should be 18 x²

product of -3x , -6x = 18x² and Sum of -3x and -6x is -9x

So, split the middle term as -3x and -6x

$9x {}^{2} - 9x + 2 = 0$

$9x {}^{2} - 3x - 6x + 2 = 0$

$3x(3x - 1) - 2(3x - 1) = 0$

$(3x - 1)(3x - 2) = 0$

Equating to 0

$3x - 1 = 0$

$3x = 1$

$x = \dfrac{1}{3}$

$3x - 2 = 0$

$3x = 2$

$x = \dfrac{2}{3}$

So, the roots are 1/3, 2/3

$so \: \alpha = \dfrac{1}{3} \: \: \beta = \dfrac{2}{3}$

Now, finding

$\dfrac{1}{ \alpha } = \dfrac{1}{ \dfrac{1}{3} }$

$\dfrac{1}{ \alpha } = 3$

$\dfrac{1}{ \beta } = \dfrac{1}{ \dfrac{2}{3} }$

$\dfrac{1}{ \beta } =\dfrac{3}{2}$

So,of 1/α and 1/β = 3, 3/2