Question

Quadratic Equation

Solve :

${9x}^{2} - 33 \sqrt{3} x + 84 = 0$

​

Answer 1

$\sf{{9x}^{2} - 33 \sqrt{3} x + 84 = 0}$

$\sf\red{According\:to\:the\:quadratic\:formula,\:the\:roots\:of }$$\sf\red{the\:equation\:are\::}$

$\sf{x \: = \: \frac{-b \: ± \sqrt{ {b}^{2} - 4ac } }{2a} }$

$\sf\red{Here,}$

$\sf\blue{a\: =\: 9}$

$\sf\blue{b\: =\: (-33√3)}$

$\sf\blue{c\: =\: 84}$

$\sf{x \: = \: \frac{-(-33√3) \: ± \:\sqrt{ {(-33√3)}^{2} - 4×9×84 } }{2×9} }$

$\sf{x \: = \: \frac{33√3 \: ± \:\sqrt{ 3267 - 3024 } }{18} }$

$\sf{x \: = \: \frac{33√3 \: ± \:\sqrt{243 } }{18} }$

$\sf{x \: = \: \frac{33√3 \: ± \:9\sqrt{3} }{18} }$

$\sf{x \: = \: \frac{33√3 \: + \:9\sqrt{3} }{18} = \frac{42√3}{18} }$

$\sf{x \: = \: \frac{33√3 \: - \:9\sqrt{3} }{18} = \frac{24√3}{18} }$

$\tt{So,Roots\:of\:the\:equation\:are =\: \frac{24√3}{18} \: \frac{42√3}{18}}$