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# Using the Quadratic Formula to find Complex Roots

Sometimes, quadratic equations have no solutions that can be expressed using real numbers. However, it is possible to use complex numbers to express these non-real solutions.
Rule

The quadratic formula is $x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a},$

where $a, b,$ and $c$ correspond with the values of a quadratic equation written in standard form, $ax^2+bx+c=0.$ It is derived by completing the square on the general standard form equation. The quadratic formula can be used to find solution(s) to quadratic equations.
Concept

## The Number of Solutions of a Quadratic Equation

The solutions of a quadratic equation in the form $ax^2+bx+c=0$ can be interpreted graphically as the zeros of the quadratic function $y=ax^2+bx+c.$

If the function has two zeros, the equation $ax^2+bx+c=0$ has two solutions, and if the function has one zero, the equation has one solution. If the function doesn't have any zeros, the equation is said to have no real solutions.
Concept

## Discriminant

In the quadratic formula, the term under the radical sign is called the discriminant.

It's possible to use the discriminant to determine the number of solutions a quadratic equation has. \begin{aligned} \mathbf{b^2-4ac} & \mathbf{\ > 0} \quad \Leftrightarrow \quad 2 \ \text{real solutions} \\ \mathbf{b^2-4ac} & \mathbf{\ = 0} \quad \Leftrightarrow \quad 1 \ \text{real solution} \\ \mathbf{b^2-4ac} & \mathbf{\ < 0} \quad \Leftrightarrow \quad 0 \ \text{real solutions} \end{aligned} The solutions to a quadratic equation correspond to the zeros of the parabola.

Method

## Complex Solutions of a Quadratic Equation

When the discriminant is negative, there are no real solutions of the quadratic equation. However, the square root of a negative number can be written as an imaginary number. This way, complex solutions of a quadratic equation can be found.
Exercise

Solve the equation using the quadratic formula. $2x^2 - 4x + 10 = 0$

Solution
Since the quadratic equation is given in standard form, we can immediately identify the constants $a=2, \quad b=\text{-} 4, \quad \text{and} \quad c=10.$ Let's substitute these values into the quadratic formula to solve the equation.
$2x^2-4x+10=0$
$x=\dfrac{\text{-} ({\color{#009600}{\text{-}4}})\pm\sqrt{({\color{#009600}{\text{-}4}})^2 -4\cdot{\color{#0000FF}{2}}\cdot{\color{#FF0000}{10}}}}{2\cdot{\color{#0000FF}{2}}}$
$x=\dfrac{4\pm\sqrt{(\text{-}4)^2 -4\cdot2\cdot10}}{2\cdot2}$
$x=\dfrac{4\pm\sqrt{16-80}}{4}$
$x=\dfrac{4\pm\sqrt{\text{-}64}}{4}$
Notice that there is a negative number inside the square root. Thus, the equation has no real solutions. However, we can calculate the complex roots using the identity $\sqrt{\text{-} a}= \sqrt{a}\cdot i.$
$x=\dfrac{4\pm\sqrt{\text{-}64}}{4}$
$x=\dfrac{4 \pm \sqrt{64} \cdot i}{4}$
$x=\dfrac{4 \pm 8i}{4}$
$x=1 \pm 2i$
The solutions to the equation are $x=1+2i$ and $x=1-2i.$
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