Question

find the solution of the quadratic equation
$3 \sqrt{3} {x}^{2} + 10x + \sqrt{3} = 0$
Answer is
$x = - \sqrt{3} \: and \: x = \frac{ - 1}{3 \sqrt{3} }$
explain it ​

Answer 1

Step-by-step explanation:

to solve 3√3x² + 10x + √3 = 0

a = 3√3, b = 10, c = √3

formula used : x = [- b ± √(b²-4ac)] / 2a

=> x = [- 10 ± √(10² - 4*3√3*√3)] / 2*3√3

=> x = [- 10 ± √ (100 - 36 )] / 2*3√3

=> x1 = (- 10 + √64) / 2*3√3

=> x2 = (- 10 - √64) / 2*3√3

=> X1 = (-10 + 8) / 6√3 => -2/6√3 = - 1/3√3

=> x2 = (- 10 - 8) / 6√3 = -18 / 6√3 => - 3/√3

therefore solutions are

x = - 1 / 3√3 and

x = - 3 / √3 * √3/√3 => - √3

[ rationalise the denominator ]

Answer 2

Step-by-step explanation:

Solution:-

$\boxed{3 \sqrt{3} {x}^{2} + 10x + \sqrt{3} = 0}$

use quadratic fromula :-

$\implies{ \boxed{ \large \blue{x = \frac{ - b \pm \sqrt{b {}^{2} - 4ac } }{2a} }}}$

by using this formula we will geta

$x = \frac{ - 10 \pm \sqrt{(10) {}^{2} - 4 \times 3 \sqrt{3} \times \sqrt{3} } }{2 \times 3 \sqrt{3} } \\ x = \frac{ - 10 \pm \sqrt{100 - 36} }{6 \sqrt{3} } \\ x = \frac{ - 10 \pm \sqrt{64} }{6 \sqrt{ 3} } \\ x = \frac{ - 10 + 8}{6 \sqrt{3} } \: \: \: and \: \: x = \frac{ - 10 - 8}{6 \sqrt{3} } \\ x = \frac{ - 2}{6 \sqrt{3} } \: \: \: and \: \: x = \frac{ - 18}{6 \sqrt{3} } \\ x = \frac{ - 1}{3 \sqrt{3} } \: \: and \: \: x = \frac{ - 3}{ \sqrt{3} }$

$\boxed{ \huge \red{x = \frac{ - 1}{3 \sqrt{3} } \: \: and \: \: x = \frac{ - 3}{ \sqrt{3} } }}$

:-note =after rationalization you will found root3