Question
Solve the following quadratic equation using formula method only $3a^2x^2 +8abx + 4b^2 = 0, a \neq 0$
✓
Answer
$3 a^2 x^2+8 a b x+4 b^2=0$
$x ^2+\frac{8 b }{3 a } x +\frac{4 b ^2}{3 a ^2}=0$
$a=1 ; b=\frac{8 b}{3 a} ; c=\frac{4 b^2}{3 a^2}$
$D=b^2-4 a c$
$=\left(\frac{8 b }{3 a }\right)^2-4(1)\left(\frac{4 b ^2}{3 a ^2}\right)$
$=\frac{64 b^2}{9 a^2}-\frac{16 b^2}{3 a^2}$
$=\frac{64 b^2-48 b^2}{9 a^2}=\frac{16 b^2}{9 a^2}$
$x =\frac{- b \pm \sqrt{ b ^2-4 ac }}{2 a}$
$x=\frac{-\frac{8 b }{3 a } \pm \sqrt{\frac{16 b ^2}{9 a ^2}}}{2}$
$x=\frac{-\frac{8 b}{3 a}+\frac{4 b}{3 a}}{2}, x=\frac{-\frac{8 b}{3 a}-\frac{4 b}{3 a}}{2}$
$x=\frac{-4 b}{6 a}, x=\frac{-12 b}{6 a}$
$x=-\frac{2 b}{3 a}, x=-\frac{2 b}{a}$
$x ^2+\frac{8 b }{3 a } x +\frac{4 b ^2}{3 a ^2}=0$
$a=1 ; b=\frac{8 b}{3 a} ; c=\frac{4 b^2}{3 a^2}$
$D=b^2-4 a c$
$=\left(\frac{8 b }{3 a }\right)^2-4(1)\left(\frac{4 b ^2}{3 a ^2}\right)$
$=\frac{64 b^2}{9 a^2}-\frac{16 b^2}{3 a^2}$
$=\frac{64 b^2-48 b^2}{9 a^2}=\frac{16 b^2}{9 a^2}$
$x =\frac{- b \pm \sqrt{ b ^2-4 ac }}{2 a}$
$x=\frac{-\frac{8 b }{3 a } \pm \sqrt{\frac{16 b ^2}{9 a ^2}}}{2}$
$x=\frac{-\frac{8 b}{3 a}+\frac{4 b}{3 a}}{2}, x=\frac{-\frac{8 b}{3 a}-\frac{4 b}{3 a}}{2}$
$x=\frac{-4 b}{6 a}, x=\frac{-12 b}{6 a}$
$x=-\frac{2 b}{3 a}, x=-\frac{2 b}{a}$
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