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Q: 8      \small 27x^2-10x+1=0.

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Given equation is
\small 27x^2-10x+1=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=27,b=-10 \ and \ c= 1
Therefore,
\frac{-(-10)\pm \sqrt{(-10)^2-4.27.1}}{2.27}= \frac{10\pm\sqrt{100-108}}{54} = \frac{10\pm\sqrt{-8}}{54}=\frac{10\pm i2\sqrt2}{54}=\frac{5}{27}\pm i\frac{\sqrt2}{27}
Therefore, the solutions of requires equation are    \frac{5}{27}\pm i\frac{\sqrt2}{27}

Posted by

Gautam harsolia

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