Question

Find the roots of the quadratic equation 3x2-2√6x+2=0 by quadratic formula

Answer 1

Step-by-step explanation:

by quadratic equations

a=3,b=-2√6,c=2

D=b^2-4ac=(-2√6)^2-4.3.2=24-24=0

x=-b±√D/2a

= -(-2√6) ±√0/2.3

2√6/6=√6/3

u can again simpilify it

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Answer 2

Roots are $x=\dfrac{1}{3}\sqrt{6},\dfrac{1}{3}\sqrt{6}$

Tips:

If $ax^{2}+bx+c=0$ be a given equation, then its roots are given by

$x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}$

Step-by-step explanation:

The given quadratic equation is

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

Comparing this with the general quadratic equation, we have

$a=3,b=-2\sqrt{6},c=2$

Now, putting these values in quadratic formula, we have

$x=\dfrac{-(-2\sqrt{6})\pm \sqrt{(-2\sqrt{6})^{2}-4\times 3\times 2}}{2\times 3}$

$=\dfrac{2\sqrt{6}\pm\sqrt{24-24}}{6}$

$=\dfrac{2\sqrt{6}\pm 0}{6}$

$=\dfrac{1}{3}\sqrt{6},\dfrac{1}{3}\sqrt{6}$

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