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College Algebra/Complex Numbers/Complex Solutions of Quadratic Equations

Complex Numbers

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Complex Solutions of Quadratic Equations

Solving Equations without a Linear Term

Quadratic equations with no linear term can be solved by isolating x2x^2 and taking the square root of both sides of the equation. If x2x^2 is equal to a negative number, then the solutions of the equation are pure imaginary numbers.
A quadratic equation is an equation, where aa, bb, and cc are real numbers and a≠0a\neq0, written in the form:
ax2+bx+c=0ax^2 + bx + c = 0
Solving an equation of the form ax2=−cax^2 = -c combines the processes of solving a quadratic equation and simplifying imaginary numbers.
Step-By-Step Example
Solving a Quadratic Equation of the Form ax2=−cax^2 = -c
Solve the equation:
5x2=−805x^2=-80
Step 1
Divide both sides of the equation by the coefficient of x2x^2.
5x25=−805x2=−16\begin{aligned}\frac {5x^2}{5}&=\frac{-80}{5}\\x^2&=-16\end{aligned}
Step 2
Take the square root of both sides of the equation.
(x2)=−16x=±−16\begin{aligned}\sqrt{(x^2)}&=\sqrt{-16}\\x&=\pm \sqrt{-16}\end{aligned}
Solution
Simplify the negative square root.
x=±−16=±−1⋅16=±(−1⋅16)=±(i⋅4)=±4i\begin{aligned}x&=\pm \sqrt{-16}\\&=\pm \sqrt{-1\cdot 16}\\&=\pm (\sqrt{-1}\cdot \sqrt{16})\\&=\pm (i\cdot4)\\&=\pm 4i\end{aligned}
The solutions are x=4ix=4i or x=−4ix=-4i.

Solving by Completing the Square

Quadratic equations with no real roots can be solved by completing the square.

The roots of a quadratic equation are its solutions. A real root is a solution of an equation that is also a real number. Some quadratic equations do not have any real roots.

One way to tell that an equation of the form ax2+bx+c=0ax^2+bx+c=0 has no real roots is by looking at the graph of the related quadratic function:
f(x)=ax2+bx+cf(x)=ax^2+bx+c
If the related function has no real zeros, or values of xx that make f(x)f(x) equal to zero, then the quadratic equation has no real roots.
The equation of a related function without any real zeros does not have any real roots. For example, the function f(x)=x2−6x+10f(x)=x^2-6x+10 has no real zeros, or values of xx that make f(x)f(x) equal to zero, because the graph does not intersect the xx-axis. Therefore, the related quadratic equation x2−6x+10=0x^2-6x+10=0 has no real roots.
When a quadratic equation has no real roots, it has a pair of complex roots that are complex conjugates. One strategy for solving quadratic equations with complex roots is completing the square.

Completing the square is the process of adding a constant term to a quadratic expression to form a perfect square trinomial. A perfect square trinomial is a trinomial, or polynomial with three terms, that can be written as the square of a binomial. For example, x2+8x+16x^2+8x+16 is a perfect square trinomial because its factored form is (x+4)2(x+4)^2.

To complete the square for a quadratic expression of the form x2+bxx^2+bx, add the value of:
(b2)2\left(\frac{b}{2}\right)^{2}
Step-By-Step Example
Complete the Square to Solve a Quadratic Equation with Complex Solutions
Solve the equation:
x2−6x+10=0x^2-6x+10=0
Step 1
Write the equation in the form x2+bx=cx^2+bx=c.
x2−6x=−10x^2-6x=-10
Step 2
Find the value of this formula:
(b2)2=(−62)2=(−3)2=9\begin{aligned}\left (\frac{b}{2} \right )^{2}&=\left ( \frac{-6}{2} \right )^{2}\\&=(-3)^{2}\\&=9\end{aligned}
Step 3
Add 9 to x2+bxx^2+bx to complete the square. Add 9 to the other side of the equation to keep the equation balanced.
x2−6x+9=−10+9x2−6x+9=−1\begin{aligned}x^2-6x+9&=-10+9\\x^2-6x+9&=-1\end{aligned}
Step 4
Factor the perfect square trinomial, or a polynomial with three terms that can be written as the square of a binomial.
(x−3)2=−1\left(x-3\right)^2=-1
Step 5
Take the square root of both sides of the equation.
x−3=±−1x−3=±i\begin{aligned}x-3&=\pm \sqrt{-1}\\x-3&=\pm i\end{aligned}
Solution
Use both the positive and negative square root to write and solve two equations.
x−3=ix=3+iorx−3=−ix=3−i\begin{aligned}x-3&=i\\x&=3+i\end{aligned}\;\;\;\;\; \text{or}\;\;\;\;\;\begin{aligned}x-3&=-i\\x&=3-i\end{aligned}
The solutions are:
x=3+iorx=3−ix=3+i\;\;\;\;\; \text{or}\;\;\;\;\;x=3-i
The solutions can be checked by substituting them into the original equation. For x=3+ix=3+i:
x2−6x+10=0(3+i)2−6(3+i)+10=0(3+i)(3+i)−6(3+i)+10=09+3i+3i+i2−18−6i+10=01+(−1)=00=0\begin{aligned}x^2-6x+10&=0\\(3+i)^2-6(3+i)+10&=0\\(3+i)(3+i)-6(3+i)+10&=0\\9+3i+3i+i^2-18-6i+10&=0\\1+(-1)&=0\\0&=0\end{aligned}
For x=3−ix=3-i:
x2−6x+10=0(3−i)2−6(3−i)+10=0(3−i)(3−i)−6(3−i)+10=09−3i−3i+i2−18+6i+10=01+(−1)=00=0\begin{aligned}x^2-6x+10&=0\\(3-i)^2-6(3-i)+10&=0\\(3-i)(3-i)-6(3-i)+10&=0\\9-3i-3i+i^2-18+6i+10&=0\\1+(-1)&=0\\0&=0\end{aligned}

Solving by Using the Quadratic Formula

Quadratic equations with no real roots can be solved by using the quadratic formula.
Another way to find solutions to quadratic equations with no real roots is by using the quadratic formula. The quadratic formula is a formula used to solve a quadratic equation of the form ax2+bx+c=0ax^2+bx+c=0 by using its coefficients:
x=−b±b2−4ac2ax=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
The quadratic formula can be used to find the solutions of any quadratic equation. Once the equation is in the standard form of a quadratic equation, ax2+bx+c=0ax^2+bx+c=0, substitute the values of aa, bb, and cc into the formula and simplify. The discriminant is the part of the quadratic formula that is under the radical sign:
b2−4acb^2-4ac
The value of the discriminant indicates the type and number of roots of the quadratic equation.
  • If the discriminant is positive, then the quadratic equation has two real roots.
  • If the discriminant is zero, then the quadratic equation has one real root.
  • If the discriminant is negative, then the quadratic equation has two complex roots, which come in conjugate pairs.
Step-By-Step Example
Using the Quadratic Formula with Complex Solutions
Solve the equation:
4x2−16x+41=04x^2-16x+41=0
Step 1
The standard form of an equation is:
ax2+bx+c=0ax^2+bx+c=0
The given equation is in standard form, where a=4a=4, b=−16b=-16, and c=41c=41. Substitute the values of aa, bb, and cc into the discriminant from the quadratic formula.
b2−4ac=(−16)2−4(4)(41)=−400\begin{aligned}b^2-4ac&=(-16)^2-4(4)(41)\\&=-400\end{aligned}
The discriminant is negative. So the equation has two complex solutions.
Step 2
Substitute the values of aa, bb, and the discriminant into the quadratic formula.
x=−b±b2−4ac2ax=−(−16)±−4002(4)\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x&=\frac{-(-16)\pm\sqrt{-400}}{2(4)}\end{aligned}
Solution
Simplify the expression.
x=16±−4008=16±20i8=2±52i\begin{aligned}x&=\frac{16\pm\sqrt{-400}}{8}\\&=\frac{16\pm 20i}{8}\\&=2\pm\frac{5}{2}i\end{aligned}
The solutions are:
x=2+52iorx=2−52ix=2+\frac{5}{2}i\;\;\;\;\;\text{or}\;\;\;\;\;x=2-\frac{5}{2}i
<Operations with Complex Numbers