Solve the equation 27 x^{2}-10 x+1=0. | vedclass

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Question

(A) The given quadratic equation is $27 x^{2}-10 x+1=0$. On comparing this equation with $a x^{2}+b x+c=0$,we obtain $a=27, b=-10$,and $c=1$. Therefor

Vedclass · Accepted Answer

$\frac{5 \pm \sqrt{2} i}{27}$

Vedclass · Answer

(A) The given quadratic equation is $27 x^{2}-10 x+1=0$. On comparing this equation with $a x^{2}+b x+c=0$,we obtain $a=27, b=-10$,and $c=1$. Therefore,the discriminant $D$ is given by $D=b^{2}-4 a c = (-10)^{2}-4(27)(1) = 100-108 = -8$. Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2 a}$,we get: $x = \frac{-(-10) \pm \sqrt{-8}}{2(27)} = \frac{10 \pm 2 \sqrt{2} i}{54}$. Dividing the numerator and denominator by $2$,we obtain $x = \frac{5 \pm \sqrt{2} i}{27}$.

Solve the equation $27 x^{2}-10 x+1=0$.

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