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2. Solving Equations
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United States of America
Grades 9-12

2.04 Quadratic equations with complex solutions

Lesson
Practice

Introduction

In Algebra 1 lesson  11.04 Solving quadratic equations using the quadratic formula  , we used the discriminant to determine the nature of the solutions to a quadratic equation. When a quadratic equation has no real solutions, then the solutions are complex numbers. We will use the methods that we studied in lesson  1.02 Solving quadratic equations  to find the exact roots to any quadratic equation.

Solving quadratic equations with complex solutions

Exploration

In the applet below, the red parabola is the graph of x^2+4=0 graphed in the real plane, and the blue parabola is the same equation graphed in the imaginary plane. Click and drag to rotate the image and explore the solutions.

Loading interactive...
  1. What do you think the solutions to the equation are based on the red parabola?
  2. What do you think the solutions to the equation are based on the blue parabola?
  3. How do you think you could find those solutions algebraically?

Graphing to solve quadratic equations with complex solutions is not practical because the graph would need to have 3 dimensions: the horizontal plane, the vertical plane, and the imaginary plane. The imaginary plane is perpendicular to the real x-plane. So we will need to use algebraic methods for solving quadratic equations with complex solutions.

We have seen in Algebra 1 that quadratic equations of the form ax^2+bx+c=0 can have 2 real solutions, 1 real solution, or no real solutions. We used the discriminant to determine the nature of the solutions. If there are no real solutions, then the solutions are complex roots (of a quadratic).

\displaystyle x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}
\bm{b^2-4ac}
discriminant
x
y

b^2-4ac>0

The equation has two real solutions

x
y

b^2-4ac=0

The equation has one real solution

x
y

b^2-4ac<0

The equation has two complex solutions

When solving quadratic equations with real coefficients that have non-real roots, we can now find the solutions by expressing them as complex numbers, with the roots being complex conjugates.

Conjugate (of a complex number)

A number created by changing the sign of the imaginary part of a complex number.

Example:

The conjugate of a+bi is a-bi.

In Algebra 1, we also discussed several methods for solving quadratic equations which could also be used to find the complex solutions of a quadratic equation:

  • Factoring is best to use when the coefficients are relatively small numbers and when the value of the discriminant is a perfect square. We will learn how to factor quadratic equations with solutions that are complex in lesson  3.03 Polynomial identities  .
  • The square root property is best to use when the equation is in the form x^2=k or in vertex form a(x-h)^2+k=0.
  • Completing the square can be used to solve any quadratic equation, but it is easiest when a=1 and b is even.
  • The quadratic formula is the best method for all other types of quadratic equations, especially ones where the coefficients are large numbers.

Examples

Example 1

Determine the number and nature of the solutions to the following equations:

a

5x^2+2x+2=0

Worked Solution
Create a strategy

We can calculate the value of the discriminant to find the types of solutions to the equation. The discriminant is equal to b^2-4ac, and for this equation we have a=5, b=2, c=2.

Apply the idea

The discriminant is (2)^2-4(5)(2)=-36, so there are two complex solutions.

Reflect and check

If the discriminant is negative, we would need to find the square root of a negative number in the quadratic formula, which results in non-real solutions. We can see that if 4ac>b^2 the discriminant will always be negative.

b

16x^2-24x+9=0

Worked Solution
Create a strategy

The discriminant is equal to b^2-4ac, and for this equation we have a=16, b=-24, c=9.

Apply the idea

The discriminant is (-24)^2-4(16)(9)=0. This tells us there will be one real solution.

Reflect and check

If the discriminant is zero, there is only one real solution because the root is repeated.

\begin{aligned}16x^2-24x+9&=0\\(4x-3)(4x-3)&=0\\x=\frac{3}{4},x&=\frac{3}{4} \end{aligned}

It is not possible to have one real solution and one complex solution for a quadratic equation with real coefficients.

Example 2

Solve the following equations, stating your solutions in the form a \pm b i:

a

2x^{2} - 6 x + 19 = 0

Worked Solution
Create a strategy

A quadratic equation in standard form ax^2+bx+c=0 has the solutions x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}, and for this equation we have a=2, b=-6, c=19.

Apply the idea
\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\left( 2\right)\left( 19\right)}}{2\left( 2\right)}Substitute a=2, b=-6, c=19
\displaystyle =\displaystyle \frac{6\pm \sqrt{36-152}}{4}Evaluate the square and products
\displaystyle =\displaystyle \frac{6\pm \sqrt{-116}}{4}Evaluate the difference in the radicand
\displaystyle =\displaystyle \frac{6\pm \sqrt{-4}\sqrt{29}}{4}Rewrite the radicand
\displaystyle =\displaystyle \frac{6\pm 2i\sqrt{29}}{4}Evaluate the radical
\displaystyle =\displaystyle \frac{6}{4}\pm \frac{2i\sqrt{29}}{4}Rewrite as two fractions
\displaystyle =\displaystyle \dfrac{3}{2}\pm \dfrac{\sqrt{29}}{2}iSimplify the quotients
Reflect and check

We can see that the two complex solutions, \dfrac{3}{2}+\dfrac{\sqrt{29}}{2}i and \dfrac{3}{2}-\dfrac{\sqrt{29}}{2}i, are complex congugates. That is, they are of the form a+bi and a-bi.

b

4x^2+9=0

Worked Solution
Create a strategy

We can rewrite this to be in the form of x^2=\dfrac{k}{a}, then use the square root property to solve.

Apply the idea
\displaystyle 4x^2+9\displaystyle =\displaystyle 0Given equation
\displaystyle 4x^2\displaystyle =\displaystyle -9Subtract 9 from both sides
\displaystyle x^2\displaystyle =\displaystyle -\dfrac{9}{4}Divide both sides by 4
\displaystyle x\displaystyle =\displaystyle \pm\sqrt{-\dfrac{9}{4}}Square root property
\displaystyle x\displaystyle =\displaystyle \pm\dfrac{\sqrt{-9}}{\sqrt{4}}Division property of radicals
\displaystyle x\displaystyle =\displaystyle \pm \dfrac{3}{2}iEvaluate the radical
Reflect and check

The value of the discriminant with a=4,b=0,c=9 is

\begin{aligned}b^2-4ac&=0^2-4(4)(9)\\&=-144 \end{aligned}

Since the discriminant is negative, there should be 2 complex solutions.

As we found above, there are 2 complex solutions, \dfrac{3}{2}i and -\dfrac{3}{2}i, as expected.

Example 3

Consider the quadratic function p\left(x\right)=x^2-6x+16.

Find the roots of the equation p \left( x \right) = 0.

Worked Solution
Create a strategy

The roots of the equation p \left( x \right) = 0 are the same as the solutions to the equation x^2-6x+16=0. Since a=1 and b is even, it would be easy to complete the square to find the roots.

Apply the idea
1\displaystyle x^2-6x+16\displaystyle =\displaystyle 0Given equation
2\displaystyle x^2-6x\displaystyle =\displaystyle -16Subtract the constant from both sides
3\displaystyle x^2-6x+9\displaystyle =\displaystyle -16+9Complete the square
4\displaystyle (x-3)^2\displaystyle =\displaystyle -7Factor the left side, evaluate the right side
5\displaystyle x-3\displaystyle =\displaystyle \pm\sqrt{-7}Square root property
6\displaystyle x\displaystyle =\displaystyle 3\pm\sqrt{-7}Add 3 to both sides
7\displaystyle x\displaystyle =\displaystyle 3\pm \sqrt{7}\sqrt{-1}Rewrite the radicand
8\displaystyle x\displaystyle =\displaystyle 3\pm \sqrt{7}iSubstitute i=\sqrt{-1}

The roots are x=3+\sqrt{7}i and x=3-\sqrt{7}i.

Reflect and check

In step 3, we completed the square by finding half of the x-term and squaring it.

In step 6, notice we wrote the 3 in front of the \pm symbol. This is because we want the answer to be in the form a\pm bi. If we had written it at the end of the equation as shown below,

x=\pm\sqrt{-7}+3

we then could have used the commutative property

x=3\pm\sqrt{-7}

to get it into the form a\pm bi.

Example 4

Consider the equation x^2+\dfrac{3}{2}x=-2.

a

Determine the nature and number of solutions.

Worked Solution
Create a strategy

To determine the nature and number, we need to find the discriminant. But before caluclating the discriminant, we need to move all terms to the same side.

\begin{aligned}x^2+\dfrac{3}{2}x&=-2\\x^2+\dfrac{3}{2}x+2&=0 \end{aligned}

For this equation, a=1, b=\dfrac{3}{2}, c=2.

Apply the idea
\displaystyle b^2-4ac\displaystyle =\displaystyle \left(\dfrac{3}{2}\right)^2-4\left(1\right)\left(2\right)Substitute coefficients into the discriminant
\displaystyle =\displaystyle \dfrac{9}{4}-8Evaluate the multiplication
\displaystyle =\displaystyle -\dfrac{23}{4}Evaluate the subtraction

Since the discriminant is negative, the equation has 2 complex solutions.

b

Find the roots of the equation.

Worked Solution
Create a strategy

Since we already have the value of the discriminant, we can use the quadratic formula and sustitute the discriminant into the radicand.

Apply the idea
1\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}State the quadratic formula
2\displaystyle =\displaystyle \dfrac{-\left(\frac{3}{2}\right)\pm\sqrt{-\frac{23}{4}}}{2(1)}Substitute known values
3\displaystyle =\displaystyle \dfrac{-\frac{3}{2}\pm\frac{\sqrt{23}}{\sqrt{4}}i}{2}Rewrite the radicand
4\displaystyle =\displaystyle \dfrac{-\frac{3}{2}\pm\frac{\sqrt{23}}{2}i}{2}Evalute the radical
5\displaystyle =\displaystyle \dfrac{-3}{4}\pm\dfrac{\sqrt{23}}{4}iEvalute the division
Reflect and check

To get from step 4 to 5, we divided the numerator of the complex fraction by the denominator:

\begin{aligned} \left(\dfrac{3}{2} \pm \dfrac{\sqrt{23}}{2}i\right) \div \dfrac{2}{1}&= \left(\dfrac{3}{2} \pm \dfrac{\sqrt{23}}{2}i\right)\times\dfrac{1}{2}\\ &=\dfrac{3}{4} \pm\dfrac{\sqrt{23}}{4}i&\end{aligned}

Idea summary

The discriminant, b^2-4ac, can help us determine the nature and number of solutions to a quadratic equation without needing to fully solve the equation.

  • b^2-4ac>0 two real solutions

  • b^2-4ac=0 one real solution

  • b^2-4ac<0 two complex solutions

We can use the square root property, completing the square, or the quadratic formula to solve quadratic equations with complex solutions. We cannot find complex solutions by graphing.

Outcomes

N.CN.C.7

Solve quadratic equations with real coefficients that have complex solutions

N.CN.C.8 (+)

Extend polynomial identities to the complex numbers.

N.CN.C.9 (+)

Know the fundamental theorem of algebra; show that it is true for quadratic polynomials.

A.REI.B.4.B

Solve quadratic equations by inspection (e.g. For x^2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

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