Solve the following


$2 x^{2}+x+1=0$


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

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

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

$\Rightarrow \alpha=\frac{-1+\sqrt{-7}}{4} \quad$ and $\quad \beta=\frac{-1-\sqrt{-7}}{4}$

$\Rightarrow \alpha=\frac{-1+i \sqrt{7}}{4} \quad$ and $\quad \beta=\frac{-1-i \sqrt{7}}{4}$

$\Rightarrow \alpha=-\frac{1}{4}+\frac{\sqrt{7}}{4} i \quad$ and $\quad \beta=-\frac{1}{4}-\frac{\sqrt{7}}{4} i$

Hence, the roots of the equation are $\frac{-1 \pm i \sqrt{7}}{4}$.

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now