# Solve the following

Question:

$17 x^{2}-8 x+1=0$

Solution:

Given: $17 x^{2}-8 x+1=0$

Comparing the given equation with the general form of the quadratic equation $a x^{2}+b x+c=0$, we get $a=17, b=-8$ 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:

$\alpha=\frac{8+\sqrt{64-4 \times 17 \times 1}}{2 \times 17} \quad$ and $\quad \beta=\frac{8-\sqrt{64-4 \times 17 \times 1}}{2 \times 17}$

$\Rightarrow \alpha=\frac{8+\sqrt{64-68}}{34}$ and $\beta=\frac{8-\sqrt{64-68}}{34}$

$\Rightarrow \alpha=\frac{8+\sqrt{-4}}{34} \quad$ and $\quad \beta=\frac{8-\sqrt{-4}}{34}$

$\Rightarrow \alpha=\frac{8+\sqrt{4 i^{2}}}{34} \quad$ and $\quad \beta=\frac{8-\sqrt{4 i^{2}}}{34}$

$\Rightarrow \alpha=\frac{8+2 i}{34} \quad$ and $\quad \beta=\frac{8-2 i}{34}$

$\Rightarrow \alpha=\frac{4+i}{17} \quad$ and $\quad \beta=\frac{4-i}{17}$

$\Rightarrow \alpha=\frac{4}{17}+\frac{1}{17} i \quad$ and $\quad \beta=\frac{4}{17}-\frac{1}{17} i$

Hence, the roots of the equation are $\frac{4}{17} \pm \frac{1}{17} i$.