Question

1) Find the roots of the quadratic equation:
3x² - 2V6x + 2 = 0​

Answer 1

Answer:

Step-by-step explanation:

3x²-2√6x+2=0

3x²-√6x-√6x+2=0

√3x(√3x-√2)-√2(√3x-√2)=0

(√3x-√2)(√3x+√2)=0

x=-√3/√2, √2/√3

Answer 2

Step-by-step explanation:

3x² - 2√6x + 2 = 0

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

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

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

So, the roots of the equation are the values of x for which

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

Now,

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

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

So, this root is repeated twice, one for each repeated factor √3x-√2

therefore the roots of 3x² - 2√6x + 2 = 0 are

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