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Q: 6      \small 3x^2-4x+\frac{20}{3}=0

Answers (1)

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Given equation is
\small 3x^2-4x+\frac{20}{3}=0 
Now, we know that the roots of the quadratic equation are given by the formula
\frac{-b\pm \sqrt{b^2-4ac}}{2a}
In this case the value of

 a=3,b=-4 \ and \ c= \frac{20}{3}
Therefore,
\frac{-(-4)\pm \sqrt{(-4)^2-4.3.\frac{20}{3}}}{2.3}= \frac{4\pm\sqrt{16-80}}{6} = \frac{4\pm\sqrt{-64}}{6}=\frac{4\pm8i}{6}= \frac{2}{3}\pm i\frac{4}{3}
Therefore, the solutions of requires equation are

   \frac{2}{3}\pm i\frac{4}{3}

Posted by

Gautam harsolia

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