# Solve the following

Question:

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

Solution:

Given: $27 x^{2}-10 x+1=0$

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

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

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

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

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

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

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

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

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

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