Question

hii friends help me to do my homework
the zeroes of 3x² - 2x + 1/3 are real and equal justify ​

Answer 1

Step-by-step explanation:

Proof using Discrimant:

$3 { x}^{2} - 2x + \frac{1}{3} = 0$

The discrimant is

$\sqrt{ {b}^{2} - 4ac}$

B is -2, A is 3 and C is 1/3 so we get

$\sqrt{ {2}^{2} - 4( 3)( \frac{1}{3}) }$

$\sqrt{4 - 4}$

$\sqrt{0 } = 0$

Since the discrimant equal zero, there is one real distinct zero.

This zero will also have a multiplicity of 2, so the zero is technically equal to each other.

To get better at solving, let solve for zero.

$3 {x}^{2} - 2x + \frac{1}{3} = 0$

$3( {x}^{2} - \frac{2}{3} x + \frac{1}{9} )$

$3( x - \frac{1}{3} )(x - \frac{1}{3} )$

$3(x - \frac{1}{3} ) {}^{2} = 0$

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

That zero is 1/3,1/3