Question

solve quartic equation 3x2 +√21x -14=0, by completing square method

Answer 1

$3 {x}^{2} + \sqrt{21}x - 14 = 0 \\ 3 {x}^{2} + \sqrt{21} x = 14 \\ 3 {x}^{2} + (2)( \frac{ \sqrt{3} }{2} )( \sqrt{7} x) = 14 \\ ( \sqrt{3}x) {}^{2} + (2) \frac{( \sqrt{3})}{2} ( \sqrt{7} x) + { (\sqrt{7}) }^{2} = 14 + 7 \\ \\ {( \sqrt{3}x + \sqrt{7} ) }^{2} = 21 \\ \sqrt{ 3}x + \sqrt{7} = + \sqrt{21} \: or \: - \sqrt{21}$

For positive one,

$\sqrt{3} x = \sqrt{21} - \sqrt{7} \\ \\ x = \frac{ \sqrt{21} - \sqrt{7} }{ \sqrt{3} }$
Rationalizing the denominator,

$x = \frac{3 \sqrt{7} - \sqrt{21} }{3}$

Similarly,

for the -ve part,

$\sqrt{3} x + \sqrt{7} = - \sqrt{21} \\ x = - ( \frac{3 \sqrt{7} + \sqrt{21}) }{3}$