Question

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

Answer 1

Given: $17x^2-8x+1=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx+c=0$, we get $a=17,b=-8$ 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:

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

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

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

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

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

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

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

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