Question

find the zeroes of the quadratic polynomial 6x^2 - 7x + 2

Answer 1

Hi !

=> 6x² - 7x + 2 = 0

=> 6x² - 4x - 3x + 2 = 0

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

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

=> 2x - 1 = 0

=> x = 1/2

or, 3x - 2 = 0

=> x = 2/3

Since zero of this given polynomial is 1/2& 2/3

____________________

Hope this helps !!

@Rajukumar111

Answer 2

hey User !!

⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜⚜

here's your
SOLUTION :::••• •●

$= 6 {x}^{2} - 7x + 2 \\ = >6 {x}^{2} - 7x + 2 = 0 \\ = > 6 {x}^{2} - 4x - 3x + 2 = 0 \\ = > 2x(3x - 2) - 1(3x - 2) = 0 \\ = > (2x - 1)(3x - 2) = 0$
Now,
$2x - 1 = 0 \\ = > x = \frac{1}{2} \\ \\ \: \: \: \: \: \: \: \: or \\ \\ \\ 3x - 2 = 0 \\ = > x = \frac{2}{3}$

So, the zeroes of the polynomial are
$\frac{1}{2} \: and \: \frac{2}{3}$

I hope this helps you..