Using the Quadratic Formula to find Complex Roots

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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

The quadratic formula is x=-b±b24ac2a, x=\dfrac{\text{-} b\pm\sqrt{b^2-4ac}}{2a},

where a,b,a, b, and cc correspond with the values of a quadratic equation written in standard form, ax2+bx+c=0.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 ax2+bx+c=0ax^2+bx+c=0 can be interpreted graphically as the zeros of the quadratic function y=ax2+bx+c. y=ax^2+bx+c.

If the function has two zeros, the equation ax2+bx+c=0ax^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.

Diskriminant Wordlist 1586 en.svg

It's possible to use the discriminant to determine the number of solutions a quadratic equation has. b24ac >02 real solutionsb24ac =01 real solutionb24ac <00 real solutions\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.

Antal lösningar till andragradsekvation
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. 2x24x+10=0 2x^2 - 4x + 10 = 0

Solution
Since the quadratic equation is given in standard form, we can immediately identify the constants a=2,b=-4,andc=10. 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.
2x24x+10=02x^2-4x+10=0
x=-(-4)±(-4)2421022x=\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=4±(-4)2421022x=\dfrac{4\pm\sqrt{(\text{-}4)^2 -4\cdot2\cdot10}}{2\cdot2}
x=4±16804x=\dfrac{4\pm\sqrt{16-80}}{4}
x=4±-644x=\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 -a=ai. \sqrt{\text{-} a}= \sqrt{a}\cdot i.
x=4±-644x=\dfrac{4\pm\sqrt{\text{-}64}}{4}
x=4±64i4x=\dfrac{4 \pm \sqrt{64} \cdot i}{4}
x=4±8i4x=\dfrac{4 \pm 8i}{4}
x=1±2ix=1 \pm 2i
The solutions to the equation are x=1+2ix=1+2i and x=12i.x=1-2i.
Show solution Show solution

Exercises

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