Question

$27x^2-10+1=0$

Answer 1

Given: $27x^2-10x+1=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx+c=0$, we get $a=27,b=-10$ and $c=1$.

Substituting these values in $\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$, we get:

$\alpha =\frac{10+\sqrt{100-4\times 27\times 1}}{2\times 27}$ and $\beta =\frac{10-\sqrt{100-4\times 27\times 1}}{2\times 27}$

$\Rightarrow \alpha =\frac{10+\sqrt{100-108}}{54}$ and $\beta =\frac{10-\sqrt{100-108}}{54}$

$\Rightarrow \alpha =\frac{10+\sqrt{-8}}{54}$ and $\beta =\frac{10-\sqrt{-8}}{54}$

$\Rightarrow \alpha =\frac{10+\sqrt{8i^2}}{54}$ and $\beta =\frac{10-\sqrt{8i^2}}{54}$

$\Rightarrow \alpha =\frac{10+i2\sqrt{2}}{54}$ and $\beta =\frac{10-i2\sqrt{2}}{54}$

$\Rightarrow \alpha =\frac{2(5+i\sqrt{2})}{54}$ and $\beta =\frac{2(5-i\sqrt{2})}{54}$

$\Rightarrow \alpha =\frac{5}{27}+\frac{\sqrt{2}}{27}i$ and $\beta =\frac{5}{27}-\frac{\sqrt{2}}{27}i$


Hence, the roots of the equation are $\frac{5}{27}\pm \frac{\sqrt{2}}{27}i$.