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8. Quadratic Equations
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Algebra 1

8.04 Solve quadratics using the quadratic formula

Lesson
Practice

The quadratic formula

Exploration

Solve each of the following quadratic equations:

  • x^2+0.5x-3=0
  • 3x^2+5x-11=0

  1. Which method did you choose and why?

  2. Did you find it challenging to solve the problems? If so, why?

When given a quadratic equation, we can try to solve it by factoring, but not all equations are factorable. We can try to solve by graphing, but we would have to estimate the solutions if they are not integer values. We can try to solve using the square root property, but the equation needs to be in a specific form first. The only method we know so far that can solve any quadratic is completing the square.

The quadratic formula is another method that can be used to solve any quadratic equation.

Before we can use the quadratic formula, we have to rearrange the quadratic equation into the form ax^{2}+bx+c=0, where a,\,b, and c are any number and a \neq 0. Once the equation is in this form, the solutions are given by the quadratic formula:

\displaystyle x=\dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a}
\bm{a}
The coefficent of x^2
\bm{b}
The coefficent of x
\bm{c}
The constant

We say that a \neq 0 because, if it were, we wouldn't have an x^2 term and the equation wouldn't be quadratic. However, the parameter a can be any other real number, and b or c can be any real number without restriction.

Examples

Example 1

The standard form of a quadratic equation is ax^2+bx+c=0.

a

Derive the quadratic formula by solving this equation for x.

Worked Solution
Create a strategy

To solve for x, we can complete the square.

Apply the idea
\displaystyle ax^2+bx+c\displaystyle =\displaystyle 0
\displaystyle x^2+\dfrac{b}{a}x+\dfrac{c}{a}\displaystyle =\displaystyle 0Division property of equality
\displaystyle x^2+\dfrac{b}{a}x\displaystyle =\displaystyle -\dfrac{c}{a}Subtraction property of equality
\displaystyle x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2\displaystyle =\displaystyle \left(\dfrac{b}{2a}\right)^2-\dfrac{c}{a}Complete the square
\displaystyle \left(x+\dfrac{b}{2a}\right)^2\displaystyle =\displaystyle \left(\dfrac{b}{2a}\right)^2-\dfrac{c}{a}Factor the left side
\displaystyle \left(x+\dfrac{b}{2a}\right)^2\displaystyle =\displaystyle \dfrac{b^2}{4a^2}-\dfrac{c}{a}Evaluate the exponent
\displaystyle \left(x+\dfrac{b}{2a}\right)^2\displaystyle =\displaystyle \dfrac{b^2-4ac}{4a^2}Evaluate the subtraction
\displaystyle x+\dfrac{b}{2a}\displaystyle =\displaystyle \pm \sqrt{\dfrac{b^2-4ac}{4a^2}}Square root property
\displaystyle x+\dfrac{b}{2a}\displaystyle =\displaystyle \pm \dfrac{\sqrt{b^2-4ac}}{2a}Evaluate the square root
\displaystyle x\displaystyle =\displaystyle -\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}Subtraction property of equality
\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}Evaluate the addition
Reflect and check

The standard form of a quadratic equation represents any quadratic equation. Since we solved this equation for x, this formula can be used to find the solution to any quadratic equation.

b

Use the quadratic formula to solve the equation -x^2+6x-8=0

Worked Solution
Create a strategy

First, we need to make sure the equation is in standard form and equal to zero. This one already is, so we can see that a=-1, b=6, and c=-8. We can substitute these values into the quadratic formula to solve for x.

Apply the idea
\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(6\right)\pm \sqrt{\left(6\right)^2-4\left(-1\right)\left(-8\right)}}{2\left(-1\right)}Quadratic formula with a=-1, b=6, and c=-8
\displaystyle =\displaystyle \dfrac{-6\pm \sqrt{36-32}}{-2}Evaluate the exponent and multiplication
\displaystyle =\displaystyle \dfrac{-6\pm \sqrt{4}}{-2}Evaluate the subtraction
\displaystyle =\displaystyle \dfrac{-6\pm 2}{-2}Evaluate the square root

Now, we can separate this into the two answers:

x=\dfrac{-6+2}{-2} and x=\dfrac{-6-2}{-2}

When we simplify, we find the answers to be x=2 and x=4.

Reflect and check

Since the answers are rational, we could have solved the quadratic equation by factoring.

\displaystyle -x^2+6x-8\displaystyle =\displaystyle 0Given equation
\displaystyle -(x^2-6x+8)\displaystyle =\displaystyle 0Factor out -1
\displaystyle -\left(x-4\right)\left(x-2\right)\displaystyle =\displaystyle 0Factor the trinomial

Using the zero product property, we get the answers x=4 and x=2.

c

Use the quadratic formula to solve the equation 5x^2=8x+1.

Worked Solution
Create a strategy

Before using the quadratic formula, we need to get the equation in the form ax^2+bx+c=0. Then, we can correctly identify the values of a, b, and c.

Apply the idea
\displaystyle 5x^2\displaystyle =\displaystyle 8x+1Given equation
\displaystyle 5x^2-8x-1\displaystyle =\displaystyle 0Subtraction property of equality

Now we can see a=5, b=-8, c=-1.

\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\left(5\right)\left(-1\right)}}{2\left(5\right)}Quadratic formula with a=5, b=-8, and c=-1
\displaystyle =\displaystyle \dfrac{8\pm \sqrt{64+20}}{10}Evaluate the exponent and multiplication
\displaystyle =\displaystyle \dfrac{8\pm\sqrt{84}}{10}Evaluate the addition
\displaystyle =\displaystyle \dfrac{8\pm\sqrt{4}\sqrt{21}}{10}Product of radicals
\displaystyle =\displaystyle \dfrac{8\pm 2\sqrt{21}}{10}Evaluate the square root
\displaystyle =\displaystyle \dfrac{4\pm \sqrt{21}}{5}Simplify by a factor of 2

The answers are x=\dfrac{4+\sqrt{21}}{5} and x=\dfrac{4-\sqrt{21}}{5}.

Reflect and check

When the answer is irrational, then the quadratic formula or completing the square are the only methods we could use to solve the quadratic equation.

Example 2

Solve:

a

5x^2-15x+2=0

Worked Solution
Create a strategy

Rearrange the equation into the form ax^{2}+bx+c=0, then use the quadratic formula.

Apply the idea

The equation 5x^2-15x+2=0 is already in standard form so we can identify a, \text{ }b, \text{ and } c. We can see that a=5, \text{ }b=-15, \text{ and } c=2 and substitute them into the quadratic formla.

\displaystyle x\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-(-15)\pm\sqrt{(-15)^{2}-4\cdot 5 \cdot 2}}{2\cdot 5}Substitute a=5,\, b=-15, \,c=2
\displaystyle x\displaystyle =\displaystyle \dfrac{15\pm\sqrt{(-15)^{2}-4\cdot 5 \cdot 2}}{2\cdot 5}Simplify the adjacent signs
\displaystyle x\displaystyle =\displaystyle \dfrac{15\pm\sqrt{(-15)^{2}-40}}{10}Evaluate the multiplication
\displaystyle x\displaystyle =\displaystyle \dfrac{15\pm\sqrt{225-40}}{10}Evaluate the exponent
\displaystyle x\displaystyle =\displaystyle \dfrac{15\pm\sqrt{185}}{10}Evaluate the subtraction

So the solutions are x=\dfrac{15+\sqrt{185}}{10} and x=\dfrac{15-\sqrt{185}}{10}.

b

10-6m+2m^{2}=m^{2}+8m+9

Worked Solution
Create a strategy

Rearrange the equation into the form ax^{2}+bx+c=0, and use the quadratic formula.

Apply the idea
\displaystyle 10-6m+2m^{2}\displaystyle =\displaystyle m^{2}+8m+9Original equation
\displaystyle 10-6m+m^{2}\displaystyle =\displaystyle 8m+9Subtract m^2 from both sides
\displaystyle 10-14m+m^{2}\displaystyle =\displaystyle 9Subtract 8m from both sides
\displaystyle 1-14m+m^{2}\displaystyle =\displaystyle 0Subtract 9 from both sides
\displaystyle m^{2}-14m+1\displaystyle =\displaystyle 0Write in descending order

Now we can see that a=1, \text{ }b=-14, \text{ and }c=1 so we can substitute these values into the quadratic formula.

\displaystyle m\displaystyle =\displaystyle \dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}Quadratic formula
\displaystyle m\displaystyle =\displaystyle \dfrac{-(-14)\pm\sqrt{(-14)^{2}-4\cdot 1 \cdot 1}}{2\cdot 1}Substitute a=1,\, b=-14, \,c=1
\displaystyle m\displaystyle =\displaystyle \dfrac{14\pm\sqrt{(-14)^{2}-4\cdot 1 \cdot 1}}{2\cdot 1}Simplify the adjacent signs
\displaystyle m\displaystyle =\displaystyle \dfrac{14\pm\sqrt{(-14)^{2}-4}}{2}Evaluate the multiplication
\displaystyle m\displaystyle =\displaystyle \dfrac{14\pm\sqrt{196-4}}{2}Evaluate the exponent
\displaystyle m\displaystyle =\displaystyle \dfrac{14\pm\sqrt{192}}{2}Simplify the expression inside the square root
\displaystyle m\displaystyle =\displaystyle \dfrac{14\pm 8\sqrt{3}}{2}Simplify the square root
\displaystyle m\displaystyle =\displaystyle 7\pm 4\sqrt{3}Divide out the common factor of 2

So the solutions are m=7+4\sqrt{3} and m=7-4\sqrt{3}.

Example 3

A ball is launched from a height of 80\text{ ft} with an initial velocity of 107\text{ ft} per second. Its height, h feet, after x seconds is given by h=-16x^2+107x+80Determine the number of seconds it will take the ball to reach the ground. Explain your reasoning.

Worked Solution
Create a strategy

On the ground the ball will have a height of 0\text{ ft}, so h=0.This means we want to solve the equation 0=-16x^2+107x+80. Since the numbers are large, the quadratic formula is an appropriate method for solving.

Apply the idea

For this equation, a=-16, b=107, and c=80. Substituting into the quadratic equation, we get

\displaystyle x\displaystyle =\displaystyle \dfrac{-(107) \pm \sqrt{\left(107\right)^2-4\left(-16\right)\left(80\right)}}{2\left(-16\right)}Quadratic formula with a=-16, b=107, and c=80
\displaystyle =\displaystyle \dfrac{-107\pm \sqrt{11449+5120}}{-32}Evaluate the exponent and multiplication
\displaystyle =\displaystyle \dfrac{-107\pm \sqrt{16569}}{-32}Evaluate the addition
\displaystyle =\displaystyle \dfrac{-107\pm \sqrt{9}\sqrt{1841}}{-32}Product of radicals
\displaystyle =\displaystyle \dfrac{-107 \pm 3\sqrt{1841}}{-32}Evaluate the square root

Therefore, x=\dfrac{-107 + 3\sqrt{1841}}{-32} and x=\dfrac{-107-3\sqrt{1841}}{-32}.

The two solutions, rounded to two decimal places, are x=-0.68 and x=7.37.Remember that x represents seconds. We can exclude the negative solution as it is outside the domain, which is x\geq 0, since time cannot be negative.

The ball will reach the ground after 7.37 seconds.

Reflect and check

In many real-world situations, negative numbers do not make sense. Always check that your answers satisfy the constraints of the variables.

Example 4

The amount of litter in a park at the end of the day can be modeled against the number of people who visited the park that day by the equation:L=-\frac{1}{50}\left(P^2-73P-150\right) where L is the number of pieces of litter and P is the number of people.

Determine the number of people who visited the park if there are 20 pieces of litter at the end of the day.

Worked Solution
Create a strategy

We want to find the number of people, P, when there are 20 pieces of litter at the end of the day, L=20.

We can do this by substituting L=20 into the equation, rearranging the equation into quadratic standard form, and then using the quadratic formula to solve for P.

Apply the idea
\displaystyle L\displaystyle =\displaystyle -\frac{1}{50}\left(P^2-73P-150\right)Model equation
\displaystyle 20\displaystyle =\displaystyle -\frac{1}{50}\left(P^2-73P-150\right)Substitute in L=20
\displaystyle -1000\displaystyle =\displaystyle P^2-73P-150Multiply both sides by -50
\displaystyle 0\displaystyle =\displaystyle P^2-73P+850Add 1000 to both sides
\displaystyle P\displaystyle =\displaystyle \frac{-(-73)\pm\sqrt{(-73)^2-4(1)(850)}}{2(1)}Quadratic formula with a=1, b=-73, and c=850
\displaystyle P\displaystyle =\displaystyle \dfrac{73\pm \sqrt{1929}}{2}Evaluate the operations

The two solutions (rounded to two decimal places) are P=58.46 and P=14.54. Since we are counting the number of people who visited the park, we want to round to the nearest whole number.

If there are 20 pieces of litter in the park at the end of the day, then either 58 or 15 people visited the park that day.

Reflect and check

In real life applications where we are counting whole objects, we want to round to the nearest integer, so our solution makes sense in context.

Idea summary

For any quadratic equation of the form 0=ax^2+bx+c where a \neq 0 and a, b, and c are real numbers, the quadratic formula can be used to solve for x.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

The discriminant

The radicand in the quadratic formula is called the discriminant.

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

The discriminant can be used to determine the number and type of solutions to any quadratic equation.

Exploration

Move the sliders to explore the questions below.

Loading interactive...
  1. What do each of the sliders represent?
  2. How many types of roots can there be?
  3. How do the types of roots relate to the value of the discriminant, D?
  4. How do the x-intercepts relate to the value of the discriminant, D?

The values of a, b, and c affect value of the discriminant changes. The type of number the discriminant affects the number and type of x-intercepts on the graph.

Quadratic equations can have 3 types of solutions: 2 real solutions, 1 real solution, or no real solutions. The value of the discriminant quickly reveals which type of solution a quadratic equation has.

-3
-2
-1
1
2
3
4
5
x
1
2
3
4
5
6
7
8
9
10
11
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13
y

Discriminant \left(>0\right):

b^2-4ac=\left(2\right)^2-4\left(-1\right)\left(12\right)=52

Solutions:

x=\dfrac{-2\pm\sqrt{52}}{-2}=1\pm\sqrt{13}

x-intercepts:

\left(1+\sqrt{13},0\right) and \left(1-\sqrt{13},0\right)

The square root of a positive number is a real number, so the plus or minus sign ensures there will always be 2 real solutions when the discriminant is positive.

-1
1
2
3
x
1
2
3
4
y

Discriminant \left(=0\right):

b^2-4ac=\left(-2\right)^2-4\left(1\right)\left(1\right)=0

Solution:

x=\dfrac{2\pm\sqrt{0}}{2}=1

x-intercept:

\left(1,0\right)

The square root of zero is zero. This eliminates the radical part of the quadratic equation, leaving only x=\dfrac{-b}{2a} which will result in a single value. So when the discriminant is zero, there will be one real solution.

-1
1
2
3
x
1
2
3
4
y

Discriminant \left(<0\right):

b^2-4ac=\left(-2\right)^2-4\left(1\right)\left(2\right)=-4

Solutions:

x=\dfrac{2\pm\sqrt{-4}}{2}

x-intercepts:

None

No real number gives us a negative number when squared. Therefore, there are no real solutions when the discriminant is negative.

Examples

Example 5

Use the discriminant to determine the number and nature of the solutions of the following quadratic equations:

a

2x^2-8x+3=0

Worked Solution
Create a strategy

For this equation, we have a=2, b=-8, c=3.

Apply the idea

The discriminant is (-8)^2-4(2)(3)=40. Since it is positive, the equation has two real solutions.

Reflect and check

Two real solutions means the function has two x-intercepts.

b

-5x^2+6x-2=0

Worked Solution
Create a strategy

For this equation, we have a=-5, b=6, c=-2.

Apply the idea

The discriminant has a value of \left(6\right)^2-4\left(-5\right)\left(-2\right)=-4. Since it is negative, the equation has no real solutions.

Reflect and check

If the discriminant is negative, then the formula will involve taking the square root of a negative number which will result in no real solutions. This quadratic will not intercept the x-axis.

We can also see that if 4ac>b^2 the discriminant will be negative, and the corresponding equation will have no real solutions.

c

x^2-3x+9=3x

Worked Solution
Create a strategy

Before identifying our variables, the equation must be in the form ax^2+bx+c=0, so we need to begin by subtracting 3x from both sides.

x^2-6x+9=0

For this equation, we have a=1, b=-6, c=9.

Apply the idea

The discriminant has a value of \left(-6\right)^2-4\left(1\right)\left(9\right)=0, so the equation has one real solution.

Reflect and check

When the discriminant is zero, the quadratic equation simplifies to be x=-\dfrac{b}{2a}, which is equal to the x-value of the vertex. This means the vertex lies on the x-axis and the x-coordinate of the vertex is the only solution to the quadratic equation.

Idea summary

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

  • b^2-4ac>0 two real solutions

  • b^2-4ac=0 one real solution

  • b^2-4ac<0 no real solutions

Outcomes

A.EI.3

The student will represent, solve, and interpret the solution to a quadratic equation in one variable.

A.EI.3a

Solve a quadratic equation in one variable over the set of real numbers with rational or irrational solutions, including those that can be used to solve contextual problems.

A.EI.3b

Determine and justify if a quadratic equation in one variable has no real solutions, one real solution, or two real solutions.

A.EI.3c

Verify possible solution(s) to a quadratic equation in one variable algebraically, graphically, and with technology to justify the reasonableness of answer(s). Explain the solution method and interpret solutions for problems given in context.

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