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

Quadratic Formula

To solve a quadratic equation written in standard form ax2+bx+c=0, the Quadratic Formula can be used.

In this formula, the discriminant b2−4ac determines the number of real solutions of the quadratic equation.

Proof

The Quadratic Formula can be derived by completing the square given the standard form of the quadratic equation ax2+bx+c=0. This method will be used to isolate the x-variable. To complete the square, there are five steps to follow.

1

Factor Out the Coefficient of x2

It is easier to complete the square when the quadratic expression is written in the form x2+bx+c. Therefore, the coefficient a should be factored out.

Since the equation is quadratic, the coefficient a is not equal to 0. Therefore, both sides of the equation can be divided by a.

2

Identify the Constant Needed to Complete the Square

The next step is to rewrite the equation by moving the existing constant to the right-hand side. To do so,

\frac{c}{a}

will be subtracted from both sides of the equation.

The constant needed to complete the square can now be identified by focusing on the x-term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the x-term, which in this case is

\frac{b}{a} .

(\frac{\frac{b}{a}}{2})^{2}

DivFracD

$\frac{a / c}{b} = \frac{a}{b \cdot c}$

(\frac{b}{2 a})^{2}

Note that leaving the constant as a power makes the next steps easier to perform.

3

Complete the Square

The square can now be completed by adding the constant found in Step 2 to both sides of the equation.

x^{2} + \frac{b}{a} x = - \frac{c}{a} ⇕ Perfect Square Trinomial x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = Constant - \frac{c}{a} + (\frac{b}{2 a})^{2}

The first three terms form a perfect square trinomial, which can be factored as the square of a binomial. The other two terms do not contain the variable x. Therefore, their value is constant.

4

Factor the Perfect Square Trinomial

The perfect square trinomial can now be factored and rewritten as the square of a binomial.

x^{2} + \frac{b}{a} x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

x^{2} + 2 (\frac{b}{2 a}) x + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

CommutativePropMult

Commutative Property of Multiplication

x^{2} + 2 x (\frac{b}{2 a}) + (\frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

FacPosPerfectSquare

a2+2ab+b2=(a+b)2

(x + \frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

The process of completing the square is now finished.

5

Simplify the Equation

Finally, the right-hand side of the equation can be simplified.

(x + \frac{b}{2 a})^{2} = - \frac{c}{a} + (\frac{b}{2 a})^{2}

Simplify right-hand side

CommutativePropAdd

Commutative Property of Addition

(x + \frac{b}{2 a})^{2} = (\frac{b}{2 a})^{2} - \frac{c}{a}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{( 2 a ) ^{2}} - \frac{c}{a}

PowProdII

$(a b)^{m} = a^{m} b^{m}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c}{a}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 4 a}{b \cdot 4 a}$

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{c \cdot 4 a}{a \cdot 4 a}

CommutativePropMult

Commutative Property of Multiplication

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{4 a c}{4 a \cdot a}

ProdToPowTwoFac

a⋅a=a2

(x + \frac{b}{2 a})^{2} = \frac{b ^{2}}{4 a ^{2}} - \frac{4 a c}{4 a ^{2}}

SubFrac

Subtract fractions

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

Now, there is only one x-term. To isolate x, it is necessary to take square roots on both sides of the equation. This results in both a positive and a negative term on the right-hand side.

Now, the equation can be further simplified to isolate x.

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}}

Solve for x

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{4 a ^{2}}

SqrtProd

$a \cdot b = a \cdot b$

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a ^{2}}

SqrtPowToNumber

$a^{2} = a$

x + \frac{b}{2 a} = \pm \frac{b ^{2} - 4 a c}{2 a}

SubEqn

$LHS - \frac{b}{2 a} = RHS - \frac{b}{2 a}$

x = - \frac{b}{2 a} \pm \frac{b ^{2} - 4 a c}{2 a}

MoveNegFracToNum

Put minus sign in numerator

x = \frac{- b}{2 a} \pm \frac{b ^{2} - 4 a c}{2 a}

AddSubFrac

Add and subtract fractions

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Finally, the Quadratic Formula has been obtained.

Concept

Number of Solutions of a Quadratic Equation

The solutions of a quadratic equation can be interpreted graphically as the zeros of the related quadratic function. Therefore, the number of solutions of a quadratic equation is the same as the number of zeros of the related function.

Quadratic Equation a x^{2} + b x + c = 0 Quadratic Function y = a x^{2} + b x + c

If the function has two zeros, the equation ax2+bx+c=0 has two solutions. Similarly, if the function has one zero, the equation has one solution. Finally, if the function does not have any zeros, the equation has no real solutions.

Concept

Discriminant

In the Quadratic Formula, the expression b2−4ac, which is under the radical symbol, is called the discriminant.

A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation.

Value of the Discriminant	Number of Real Solutions
b2−4ac>0	2
b2−4ac=0	1
b2−4ac<0	0

Moreover, the discriminant determines the number of x-intercepts of the graph of the related quadratic function.

The number of $x$-intercepts of the graph of a quadratic function

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.

Find the complex solutions of the quadratic equation

fullscreen

Solve the equation using the quadratic formula.

2x2−4x+10=0

Show Solution

Since the quadratic equation is given in standard form, we can immediately identify the constants

a = 2, b = - 4, and c = 10 .

Let's substitute these values into the quadratic formula to solve the equation.

2x2−4x+10=0

UseQuadForm

Use the Quadratic Formula: a=2,b=-4,c=10

x = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 1 0}{2 \cdot 2}

NegNeg

-(-a)=a

x = \frac{4 \pm ( - 4 ) ^{2} - 4 \cdot 2 \cdot 1 0}{2 \cdot 2}

CalcPowProd

Calculate power and product

x = \frac{4 \pm 1 6 - 8 0}{4}

SubTerm

Subtract term

x = \frac{4 \pm - 6 4}{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

x = \frac{4 \pm - 6 4}{4}

SqrtNegToSqrtI

$- a = a \cdot i$

x = \frac{4 \pm 6 4 \cdot i}{4}

CalcRoot

Calculate root

x = \frac{4 \pm 8 i}{4}

SimpQuot

Simplify quotient

x=1±2i

The solutions to the equation are x=1+2i and x=1−2i.

Channels

Direct messages

Rule

Quadratic Formula

Proof

Concept

Number of Solutions of a Quadratic Equation

Concept

Discriminant

Method

Complex Solutions of a Quadratic Equation

Example

Find the complex solutions of the quadratic equation