Question

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

Answer 1

Answer:

(√3x-√2) (√3x-√2)

Step-by-step explanation:

3x^2-2√6x+2

=(√3x) ^2 - 2(√3x) (√2) +(√2)^2

=(√3x - √2)^2

= (√3x -√2) (√3x -√2)

Answer 2

Given,

$the \: quadrtic \: polynomial \: is \: 3 {x}^{2} - 2\sqrt{6} x + 2$

To find out,

The roots of the polynomial.

Solution:

By factorisation method:

1)Here,first we have to split the middle term.

2)We have to find two numbers whose sum is -2√6 and product is 6.

3) -(√6) + -(√6) = -2√6

-(√6)×-(√6) = 6

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

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

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

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

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

$i)\sqrt{3} x - \sqrt{2} = 0 \: \: \: \: ii) \: \: \sqrt{3} x = \sqrt{2} \: \: \: \: iii)x = \frac{ \sqrt{2} }{ \sqrt{3} }$

$i) \sqrt{3} x - \sqrt{2} = 0 \: \: \: \: ii) \sqrt{3} x = \sqrt{2} \: \: \: \: \: iii)x \: = \frac{ \sqrt{2} }{ \sqrt{3} }$

$therefore \: the \: roots \: of \: the \: quadratic \: polynomial \: are \: \frac{ \sqrt{2} }{ \sqrt{3} } and \: \frac{ \sqrt{2} }{ \sqrt{3} }$