Question

$17x^2+28x+12=0$

Answer 1

Given: $17x^2+28x+12=0$
Comparing the given equation with the general form of the quadratic equation $a{x}^2+bx+c=0$, we get $a=17,b=28$ and $c=12$.


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

$\alpha =\frac{-28+\sqrt{784-4\times 17\times 12}}{34}$ and $\beta =\frac{-28-\sqrt{784-4\times 17\times 12}}{34}$

$\Rightarrow \alpha =\frac{-28+\sqrt{784-816}}{34}$ and $\beta =\frac{-28-\sqrt{784-816}}{34}$

$\Rightarrow \alpha =\frac{-28+\sqrt{-32}}{34}$ and $\beta =\frac{-28-\sqrt{-32}}{34}$

$\Rightarrow \alpha =\frac{-28+\sqrt{32i^2}}{34}$ and $\beta =\frac{-28-\sqrt{32i^2}}{34}$

$\Rightarrow \alpha =\frac{-28+4\sqrt{2}i}{34}$ and $\beta =\frac{-28-4\sqrt{2}i}{34}$

$\Rightarrow \alpha =\frac{-14+2\sqrt{2}i}{17}$ and $\beta =\frac{-14-2\sqrt{2}i}{17}$


Hence, the roots of the equation are $-\frac{14}{17}\pm \frac{2\sqrt{2}}{17}i.$