Question

Find the root of the equation $3x^2-2\sqrt6x+2=0$.

• A. $$x=\sqrt{\frac{3}{2}},\sqrt{\frac{2}{3}}$$
• B. $$x=\sqrt{\frac{2}{3}},\sqrt{-\frac{2}{3}}$$
• C. $$x=\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}$$
• D. $$x=\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}$$

Answer 1

Answer:

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

$3x^2-2\sqrt6x+2=0$

product = 3 x 2 =6

sum = -2√6

-√6 - √6 = 2√6

-√6 x - √6 = 6

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

$\sqrt 3 \times \sqrt 3x^2 - \sqrt3\sqrt2x-\sqrt3\sqrt2x+\sqrt2\sqrt2=0$

$\sqrt3x(\sqrt3x-\sqrt2)-\sqrt2(\sqrt3x-\sqrt2)=0$

$(\sqrt3x-\sqrt2)(\sqrt3x-\sqrt2)=0$

$(\sqrt3x-\sqrt2)=0$

$\sqrt3x=\sqrt2$

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

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