Question

factories the following

$2x { }^{2} - \sqrt{3x } - 3$
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Answer 1

$2 {x}^{2} - \sqrt{3} x - 3 = 0 \\ \\ \frac{2}{2} {x}^{2} - \frac{ \sqrt{3} }{2}x = \frac{3}{2} \\ \\ {(x)}^{2} - 2 \times \frac{ \sqrt{3} }{4} \times x + {\left( \frac{ \sqrt{3} }{4} \right) }^{2} = \frac{3}{2} + {\left( \frac{ \sqrt{3} }{4} \right) }^{2} \\ \\ \left( x - \frac{ \sqrt{3} }{4} \right)^{2} = \frac{3}{2} + \frac{3}{16} \\ \\ \left( x - \frac{ \sqrt{3} }{4} \right)^{2} = \frac{24 + 3}{16} \\ \\ \left( x - \frac{ \sqrt{3} }{4} \right)^{2} = \frac{27}{16} \\ \\ x - \frac{ \sqrt{3} }{4} = \pm \: \frac{3 \sqrt{3} }{4}$

Case 1

$x - \frac{ \sqrt{3} }{4} = \frac{3 \sqrt{3} }{4} \\ \\ x = \frac{ \sqrt{3} }{4} + \frac{3 \sqrt{3} }{4} \\ \\ x = \frac{ \cancel{4} \sqrt{3} }{ \cancel{4}} \\ \\ x = \sqrt{3}$

Case 2

$x - \frac{ \sqrt{3} }{4} = - \frac{ 3\sqrt{3} }{4} \\ \\ x = - \frac{3 \sqrt{3} }{4} + \frac{ \sqrt{3} }{4} \\ \\ x = \frac{ - 3 \sqrt{3} + \sqrt{3} }{4} \\ \\ x = - \frac{ \cancel{2} \sqrt{3} }{ \cancel{4}} \\ \\ x = - \frac{ \sqrt{3} }{2}$

Hence Values of x are:

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