Question

$2 \sqrt{3} x {}^{2} - \: 5x \: + \: \: \sqrt{3}$
Obtain the zeroes and find the relationship between zeroes and polynomial

Answer 1

HEY MATE

HERE IS UR ANS

REFER TO THE ATTACHMENT⤴⤴

Answer 2

Hey!!!!

Good Evening

As promised I am here to help you

Difficulty Level : Average

Chances of being asked in Board : 70%

Note : I've made a correction in the question

_________________

We have

=> 2√3x² - 5x - √3 = 0

So by middle term splitting method

Here 2√3 x √3 = 6

Thus

=> 2√3x² - 6x + x - √3 = 0

=> 2√3x(x - √3) + (x - √3) = 0

Thus

=> (2√3x + 1)(x - √3) = 0

Thus x = - 1/2√3 and x = √3

Now relationship between zeros and their co efficients


$let \: \alpha = \frac{ - 1}{2 \sqrt{3} } \: and \: \beta = \sqrt{3}$

Thus we know

$= > \alpha + \beta = \frac{ - b}{a}$

Thus

$= > \frac{ - 1}{2 \sqrt{3} } + \sqrt{3} = \frac{5}{2 \sqrt{3} }$

$= > \frac{ - 1 + 6}{2 \sqrt{3} } = \frac{5}{2 \sqrt{3} }$

Thus

$= > \frac{5}{2 \sqrt{3} } = \frac{5}{2 \sqrt{3} }$

Similarly

$= > \alpha \beta = \frac{c}{a}$

$= > \frac{ - 1}{2 \sqrt{3} } \times \sqrt{3} = \frac{ \sqrt{3} }{ - 2 \sqrt{3} }$

Thus

$= > \frac{ - 1}{2} = \frac{ - 1}{2}$
Hence Verified

__________________

Hope this helps ✌️

Good Night