Question

In each of the following, determine whether the given values are solutions of the given equation or not :

$x^2$ + x + 1 = 0, x = 0, x = 1

• A. $x=0$ and $x=1$ are solutions.
• B. $x=0$ is a solution but $x=1$ is not a solution.
• C. $x=0$ and $x=1$ are not solutions.
• D. $x=1$ is a solution but $x=0$ is not a solution.

Answer 1

The correct option is C $x=0$ and $x=1$ are not solutions.

$\Rightarrow$ Given quadratic equation is $x^2+x+1=0$, comparing it with $a{x}^2+bx+c=0$

$\Rightarrow$ We get, $a=1,b=1,c=1$

$\Rightarrow$ $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$=\frac{-1\pm \sqrt{(1{)}^2-4(1\left)\right(1)}}{2\left(1\right)}$

$=\frac{-1\pm \sqrt{1-4}}{2}$

$=\frac{-1\sqrt{-3}}{2}$

$=\frac{-1\pm \sqrt{3}i}{2}$

$\therefore$ $x=\frac{-1+\sqrt{3}i}{2}$ and $x=\frac{-1-\sqrt{3}i}{2}$

$\therefore$ Here, we can see correct answer is option C