Solve equation 2 x^{2}+x+1=0

English

Gujarati

Hindi

4-1.Complex numbers

medium

Solve the equation $2 x^{2}+x+1=0$

A

$\frac{-1 \pm \sqrt{7} i}{4}$

B

$\frac{-1 \pm \sqrt{7} i}{4}$

C

$\frac{-1 \pm \sqrt{7} i}{4}$

D

$\frac{-1 \pm \sqrt{7} i}{4}$

Solution

The given quadratic equation is $2 x^{2}+x+1=0$

On comparing the given equation with $a x^{2}+b x+c=0$

We obtain $a=2, b=1$ and $c=1$

Therefore, the discriminant of the given equation is

$D=b^{2}-4 a c=1^{2}-4 \times 2 \times 1=1-8=-7$

Therefore, the required solutions are

$\frac{-b \pm \sqrt{D}}{2 a}=\frac{-1 \pm \sqrt{-7}}{2 \times 2}=\frac{-1 \pm \sqrt{7} i}{4} \quad[\sqrt{-1}=i]$

Standard 11

Mathematics

Share

Similar Questions

Let $z$ be a complex number such that the imaginary part of $z$ is non zero and $a=z^2+z+1$ is real. Then a cannot take the value

hard

(IIT-2012)

Let $\frac{{1 – ix}}{{1 + ix}} = a – ib$ and ${a^2} + {b^2} = 1$, where $a$ and $b$ are real, then $x = $

medium

If $n$ is a positive integer, then ${\left( {\frac{{1 + i}}{{1 – i}}} \right)^{4n + 1}}$=

medium

Let ${z_1},{z_2}$ be two complex numbers such that ${z_1} + {z_2}$ and ${z_1}{z_2}$ both are real, then

medium

If $\,\left| \begin{array}{l}\,6i\,\,\,\,\, – 3i\,\,\,\,\,\,\,\,\,1\\\,\,4\,\,\,\,\,\,\,\,\,3i\,\,\,\,\,\, – 1\\\,20\,\,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\,\,i\end{array} \right|\,$=$x + iy$, then $(x, y)$ is

easy