Solving Quadratic Equations using the Quadratic Formula

The general form of a quadratic equation is $a x^{2} + b x + c = 0,$

where

a,

b,

and

c

are real numbers, and

a \neq = 0 .

There are many ways to solve such equations. One way is by using the quadratic formula, which is derived by completing the square.

Rule

The Quadratic Formula

The quadratic formula is $x = \frac{- b \pm b ^{2} - 4 a c}{2 a},$

where

a, b,

and

c

correspond with the values of a quadratic equation written in standard form,

a x^{2} + b x + 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.

Proof

Deriving the Quadratic Formula

The quadratic formula can be derived by completing the square on the general standard form a quadratic equation. Recall that completing the square is a method for solving quadratic equations. Thus, completing the square on the general form of a quadratic equation,

a x^{2} + b x + c = 0,

creates another way to solve all quadratic equations. By completing the square, it's possible to isolate

x .

The first step is to rewrite the equation by moving

c

and

a

to the right-hand side.

a x^{2} + b x + c = 0

SubEqn

LHS - c = RHS - c

a x^{2} + b x = - c

DivEqn

LHS / a = RHS / a

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

WriteSumFracWrite as a sum of fractions

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

SimpQuotSimplify quotient

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

MovePartNumRight

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

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

Next, a new constant term can be added to both sides of the equation, so the expression on the left becomes a perfect square trinomial. The coefficient of the

x

-term is

\frac{b}{a} .

Thus, the term needed to complete the square is

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

For equality to hold, this term is added on both sides.

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

AddEqn

LHS + (\frac{b}{a} / 2)^{2} = RHS + (\frac{b}{a} / 2)^{2}

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

DivFracSL

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

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

Now that the left-hand side is a perfect square trinomial, it can be written in factored form. The right-hand side can also be simplified.

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

FacPosPerfectSquare

a^{2} + 2 a b + b^{2} = (a + b)^{2}

(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{4 a c}{4 a ^{2}}

SubFracSubtract fractions

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

Now, there is one

x

-term. To isolate

x,

it is necessary to square-root both sides of the equation. This results in both a positive and a negative term on the right-hand side.

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

SqrtEqn

LHS = RHS

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

SqrtQuot

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

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}

MoveNegFracToNumPut minus sign in numerator

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

AddSubFracAdd and subtract fractions

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

Thus, it is possible to solve any quadratic equation in standard form using

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

Solve the equation using the quadratic formula

fullscreen

Exercise

Use the quadratic formula to solve the equation. $2 x^{2} - 4 x - 16 = 0$

Show Solution

Solution

Notice that the given equation is written in standard form. Thus, it can be solved using the quadratic formula. To begin, it's necessary to note the values of

a, b,

and

c .

It can be seen that

a = 2, b = - 4, and c = - 16 .

To solve the equation, we can substitute these values into the formula and simplify. Remember, since a quadratic function can have

0, 1,

or

2

roots, this equation can have

0, 1,

or

2

real solutions.

2 x^{2} - 4 x - 16 = 0

UseQuadFormUse the Quadratic Formula:

a = 2, b = - 4, c = - 16

x = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 1 6 )}{2 \cdot 2}

NegNeg

- (- a) = a

x = \frac{4 \pm ( - 4 ) ^{2} - 4 \cdot 2 ( - 1 6 )}{2 \cdot 2}

CalcPowProdCalculate power and product

x = \frac{4 \pm 1 6 + 1 2 8}{4}

AddTermsAdd terms

x = \frac{4 \pm 1 4 4}{4}

CalcRootCalculate root

x = \frac{4 \pm 1 2}{4}

StateSolState solutions

x = - 8 / 4 x = 16 / 4

CalcQuotCalculate quotient

x_{1} = - 2 x_{2} = 4

The solutions to the equation are

x = - 2

and

x = 4 .

Concept

The Number of Solutions of a Quadratic Equation

The solutions of a quadratic equation in the form $a x^{2} + b x + c = 0$ can be interpreted graphically as the zeros of the quadratic function $y = a x^{2} + b x + c .$ If the function has two zeros, the equation $a x^{2} + b x + c = 0$ 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.

two zeros

One zero

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. $b^{2} - 4 a c b^{2} - 4 a c b^{2} - 4 a c > 0 \Leftrightarrow 2 real solutions = 0 \Leftrightarrow 1 real solution < 0 \Leftrightarrow 0 real solutions$ The solutions to a quadratic equation correspond to the zeros of the parabola.

Determine the number of solutions of the quadratic equations

fullscreen

Exercise

Determine the number of real solutions to the equations without solving them. $x^{2} - 2 x + 9 = 0 and x^{2} - 4 x + 4 = 0$

Show Solution

Solution

It's possible to use the discriminant of the quadratic formula to determine the number of solutions a quadratic equation has. We'll first focus on $x^{2} - 2 x + 9 = 0 .$ Since the equation is written in standard form, we can see that $a = 1, b = - 2,$ and $c = 9 .$ We can substitute these values into the discriminant and simplify.

b^{2} - 4 a c

SubstituteValuesSubstitute values

(- 2)^{2} - 4 \cdot 1 \cdot 9

CalcPowProdCalculate power and product

4 - 36

SubTermSubtract term

- 32

Since

- 32 < 0,

the first quadratic equation has

0

real solutions. We can perform the same process on the second equation where

a = 1, b = - 4,

and

c = 4 .

b^{2} - 4 a c

SubstituteValuesSubstitute values

(- 4)^{2} - 4 \cdot 1 \cdot 4

CalcPowProdCalculate power and product

16 - 16

SubTermSubtract term

0

Since

0 = 0,

the second quadratic equation has

1

real solution.

Solving Quadratic Equations using the Quadratic Formula

The Quadratic Formula

Proof

Example

Solve the equation using the quadratic formula

The Number of Solutions of a Quadratic Equation

Discriminant

Example

Determine the number of solutions of the quadratic equations