Question

solve this quadratic equation 3 x square minus 2 root 6 X + 2 is equal to zero

Answer 1

Hi there!

Given eqn. :-
3x² - 2√6x + 2 = 0

Here, a = 3, b = -2√6 and c = 2

Discriminant = b² - 4ac

= (-2√6)² - 4(3)(2)

= 4 × 6 - 24

= 24 - 24

= 0. ---( roots are real n' equal )

by Quadratic formula :-

x = -b ± √D / 2a

x = -(-2√6) ± √0 / 2 × 3

x = 2√6 ± 0 / 6

x = 2√6 / 6

x = √6 / 3

Hence, The required answer is :- x = √6 / 3

Hope it helps!

Answer 2

Step-by-step explanation:

Given: 3$x^{2}$-$\sqrt[2]{6} x$+2 =0

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

$\sqrt{3} x$($\sqrt{3} x-\sqrt{2} x$)-$\sqrt{2}$($\sqrt{3} x-\sqrt{2} x$)=0

($\sqrt{3} x-\sqrt{2}$)($\sqrt{3} x-\sqrt{2}$)

$\sqrt{3} x$$-\sqrt{2}$=0 or $\sqrt{3} x$$-\sqrt{2}$=0

$\sqrt{3} x$= $\sqrt{2}$ or $\sqrt{3}$

$\sqrt{2}$,x= $\sqrt{\frac{2}{3} }$ or x= $\sqrt{\frac{2}{3} }$

So, the root is repeated twice, one for each repeated factor ($\sqrt{3} x$-$\sqrt{2}$)

The two equal roots of 3$x^{2}$-$\sqrt[2]{6} x$+2 =0 are  $\sqrt{\frac{2}{3} }$ , $\sqrt{\frac{2}{3} }$ .