Question

What are the solutions to the equation ${x^2} + 2x + 2 = 0$?

Answer 1

Hint: The given equation is a quadratic equation so it will have two roots. To find these roots, we are going to use the quadratic formula, that is
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
By putting the values of $a,b,c$ we will get the roots of our equation.

Complete step-by-step solution:
In this question, we have to find the solution of the equation ${x^2} + 2x + 2 = 0$.
Now, this is a quadratic equation and it will have two possible answers.
For solving this quadratic equation, we are going to use the quadratic formula.
The quadratic formula for a given quadratic equation $a{x^2} + bx + c = 0$ is given below:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ - - - - - - - - - - - - - - (1)
Where, $b$ is the coefficient of $x$and $a$ is the coefficient of ${x^2}$and c is the constant.
Our given equation is: ${x^2} + 2x + 2 = 0$
Therefore, $a = 1,b = 2,c = 2$
Putting this values in equation (1), we get
$ \Rightarrow x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times 2} }}{{2 \times 1}}$
Solving the equation, we get
$
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 8} }}{2} \\
\Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 4} }}{2} \\
 $
Now, since there is a negative number under the square root, the roots of the equation will be imaginary.
And we know that we can denote imaginary numbers with complex numbers $i$.
Therefore, above equation becomes
$
\Rightarrow x = \dfrac{{ - 2 \pm 2i}}{2} \\
\Rightarrow x = \dfrac{{ - 2 + 2i}}{2} \\
\Rightarrow x = \dfrac{{2\left( {i - 1} \right)}}{2} \\
\Rightarrow x = i - 1 \\
 $ $
\Rightarrow x = \dfrac{{ - 2 \pm 2i}}{2} \\
\Rightarrow x = \dfrac{{ - 2 - 2i}}{2} \\
\Rightarrow x = \dfrac{{2\left( { - i - 1} \right)}}{2} \\
\Rightarrow x = - i - 1 \\
 $
Therefore, the imaginary values of $x$ in equation ${x^2} + 2x + 2 = 0$ are $x = i - 1$ and $x = - i - 1$.

Note: In the quadratic formula,
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
The term ${b^2} - 4ac$ is known as the discriminant$\left( D \right)$ of the equation.
(i). If the value of $D > 0$, then the roots of the equation are real and distinct.
(ii). If the value of $D = 0$, then the equation has only one root.
(iii). If the value of $D < 0$, then the equation has imaginary roots.