Question

find the zeros of the quadratic polynomial
$\sqrt{3 }x { }^{2} - 8x + 4 \sqrt{3 }$
​

Answer 1

The two zeroes of the polynomial are

$\frac{2}{ \sqrt{3} } \: \: \: and \: \: \: 2 \sqrt{3}$

HOPE THIS COULD HELP!!!

Answer 2

Answer :-

→ x = 2/√3 or 2√3 .

Step-by-step explanation :-

We have,

→ √3x² - 8x + 4√3 = 0 .

→ √3x² - 6x - 2x + 4√3 = 0 .

→ √3x( x - 2√3 ) - 2( x - 2√3 ) = 0 .

→ ( √3x - 2 )( x - 2√3 ) = 0 .

→ √3x - 2 = 0 or x - 2√3 = 0 .

$\therefore$ x = $\frac{2}{\sqrt{3}}$ or x = $2 \sqrt{3}$ .

Hence, it is solved .