Question

Find the roots of following quadratic eqution :15x×x-10√6x+10

Answer 1

$hi$
$15 {x}^{2} - 10 \sqrt{6} x + 10 = 0$
[Dividing throughout by 5]

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

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

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

so, we get,

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

√3x - √2 = 0

√3x = √2

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


hope this helps ! !#

Answer 2

Discriminant=b^2-4ac
(10√6)^2 - 4*15*10
=600 - 600
=0..

Since discriminant is 0,it has 2 equal roots..

X = -b±√discriminant÷2a
=-(-10√6)±√0÷2×15
=10√6÷30
=√6÷3
=√2*√3÷√3*√3
=√2÷√3...