Question

Solve the quadratic: $2x^2+x+1=0$

Answer 1

We have, $2x^2+x+1=0$
On comparing with general form of quadratic equation:

$a{x}^2+bx+c=0$

We get $a=2,b=1,c=1$
Substituting these values in

$\alpha =\frac{-b+\sqrt{b^2-4ac}}{2a}and\beta =\frac{-b-\sqrt{b^2-4ac}}{2a}$

We will obtain

$\alpha =\frac{-1+\sqrt{1^2-4\times 2\times 1}}{2\times 2}$

$=\frac{-1+\sqrt{1-8}}{4}=\frac{1++\sqrt{-7}}{4}$

$\alpha =\frac{-1+\sqrt{7}\cdot \sqrt{(-1)}}{4}$

$=\frac{-1+\sqrt{7}i}{4}[\because i=\sqrt{(-1)}]$

$and\beta =\frac{-1-\sqrt{1^2-4\times 2\times 1}}{2\times 2}=\frac{-1-\sqrt{7}i}{4}$

Hence, the roots are $\frac{-1+\sqrt{7}i}{4}\&\frac{-1-\sqrt{7}i}{4}$