Question

Solve for x by quadratic 2a^2x^2+b(6a^2+1)x+3b^2=0

Answer 1

$D = b^{2} - 4ac \\ = {b(6a ^{2} + 1 )}^{2} - 4(2a^{2} ) (3b ^{2} ) \\ = (6a^{2} b + b)^{2} - 24a^{2} b ^{2} \\ = 36a ^{4} b^{2} + b^{2} + 12a ^{2} b^{2} -24a^{2} b^{2} \: \: \: \\ = 36a ^{4 }b ^{2} + b ^{2} - 12a^{2} b ^{2} = \\ = (6a^{2} b - b) ^{2} = > {using \: identitity \: (a - b) ^{2} = a ^{2} + b ^{2} - 2ab } \\ x = \frac{ - b + - \sqrt{ D } }{2a} \\ x = \frac{-6a^{2} b - b + 6a^{2} b -b}{4a ^{2} } = \frac{ - 2b}{4a ^{2} } = \frac{ - b}{2a ^{2} } \\ or \frac{ - 6a ^{2}b - b - 6a ^{2}b + b }{4a^{2} } = \frac{ - 12a ^{2}b }{4a ^{2} } = - 3b$

Answer 2

here is your answer