Question

$13x^2+7x+1=0$

Answer 1

Given: $13x^2+7x+1=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx+c=0$, we get $a=13,b=7$ 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:

$\alpha =\frac{-7+\sqrt{49-4\times 13\times 1}}{2\times 13}$ and $\beta =\frac{-7-\sqrt{49-4\times 13\times 1}}{2\times 13}$

$\Rightarrow \alpha =\frac{-7+\sqrt{49-52}}{26}$ and $\beta =\frac{-7-\sqrt{49-52}}{26}$

$\Rightarrow \alpha =\frac{-7+\sqrt{-3}}{26}$ and $\beta =\frac{-7-\sqrt{-3}}{26}$

$\Rightarrow \alpha =\frac{-7+i\sqrt{3}}{26}$ and $\beta =\frac{-7-i\sqrt{3}}{26}$

$\Rightarrow \alpha =-\frac{7}{26}+\frac{\sqrt{3}}{26}i$ and $\beta =-\frac{7}{26}-\frac{\sqrt{3}}{26}i$


Hence, the roots of the equation are $-\frac{7}{26}\pm \frac{\sqrt{3}}{26}i.$