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3. Quadratic Functions & Equations
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United States of AmericaTennessee
Algebra 2

3.02 Solutions to quadratic equations

Lesson
Practice
Lesson

Concept summary

We have several methods we can use to solve quadratic equations. To determine which method is the most suitable we need to look at the form of the quadratic equation. Some are easily solved by factoring for example, or by taking the square root. We may first need to do a process called completing the square in order to solve by taking square roots.

If we are unable to solve the quadratic easily using one of these methods, the quadratic formula is often the best approach since it can be used to solve any quadratic equation once it's written in standard form.

Quadratic formula

The formula x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}, used to find solution(s) to a quadratic equation of the form y=ax^2+bx+c

If we have access to technology, drawing the graph of the corresponding quadratic function can help us find exact solutions, or approximate a solution if it is not an integer value.

We can see in the quadratic formula that we have the term \pm\sqrt{b^2-4ac}. The expression b^2-4ac is called the discriminant, and we can use its value to determine the number of solutions to the equation.

  • If the discriminant is equal to zero, then we have one real solution: x=\dfrac{-b\pm \sqrt{0}}{2a}=\dfrac{-b}{2a}
  • If the discriminant is positive, then we have two real solutions: x=\dfrac{-b+ \sqrt{b^2-4ac}}{2a},\,x=\dfrac{-b- \sqrt{b^2-4ac}}{2a}
  • If the discriminant is negative, then \sqrt{b^2-4ac} will not be a real number, so there will be no real solutions to the equation

These cases can be distinguished visually by the location of the vertex with respect to the x-axis.

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Worked examples

Example 1

For the following quadratic equations, determine an appropriate strategy for solving, explaining your choice, and then solve for x.

a

x^2-2x-24=0

Approach

The leading coefficient of x^2 is 1, so we can check if this equation can be easily factored. The factor pairs of 24 are

  • 1\cdot24
  • 2\cdot12
  • 3\cdot8
  • 4\cdot6

and we want to find two factors that have a product of -24 and sum to -2. As the product is negative we know one factor must be negative and the other positive.

Solution

The two factors that have a product of -24 and a sum of -2 are -6 and 4. We can write the equation in factored form as \left(x-6\right)\left(x+4\right)=0, which gives us two solutions x=6,\, x=-4.

Reflection

In general, if the coefficients are small, and especially if a=1, it is worth checking to see if we can easily factor the equation to solve.

b

3\left(x-5\right)^2-27 = 0

Approach

As this is written in vertex form, we can solve it using square roots and inverse operations to solve for x.

Solution

\displaystyle 3\left(x-5\right)^2-27\displaystyle =\displaystyle 0
\displaystyle 3\left(x-5\right)^2\displaystyle =\displaystyle 27Addition property of equality
\displaystyle \left(x-5\right)^2\displaystyle =\displaystyle 9Division property of equality
\displaystyle x-5\displaystyle =\displaystyle \pm 3Take the square root of both sides
\displaystyle x\displaystyle =\displaystyle 5\pm 3 Addition property of equality

Giving us two solutions x = 2,\, x = 8

Reflection

In general, if we can easily rearrange the equation into the form \left(x-h\right)^2=k for some positive value of k then solving using square roots is a suitable method.

c

3x^2-5x+12=0

Approach

In general, if the leading coefficient is not 1 then factoring is not likely to be efficient. In this particular case the most efficient method would be to use the quadratic formula. We can also calculate the discriminant b^2-4ac, to identify if there are two, one or no real solutions.

Solution

\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-5\right) \pm \sqrt{\left(-5\right)^2-4\left(3\right)\left(12\right)}}{2\left(3\right)}Substitute values for a,b,c
\displaystyle x\displaystyle =\displaystyle \dfrac{5 \pm \sqrt{-119}}{6}Simplify

We can see that the discriminant is equal to -119. As it is less than zero, we know that the quadratic equation has no real solutions.

Reflection

Using the quadratic equation will always be an appropriate method, and has the advantage of identifying the number and type of solutions, whether they are real or non-real, or rational or non-rational. If you cannot quickly and easily identify a way to solve it using one of the other methods, then using the quadratic formula is always suitable.

d

V=\dfrac{1}{3}\pi x^2 y

Approach

When solving a literal equation that is quadratic, we can rearrange the equation to isolate the perfect square.

In this case, we want to isolate x^2 on one side of the equation.

Solution

\displaystyle V\displaystyle =\displaystyle \dfrac{1}{3}\pi x^2 yState the literal equation
\displaystyle 3V\displaystyle =\displaystyle \pi x^2 yMultiplication property of equality
\displaystyle \dfrac{3V}{\pi y}\displaystyle =\displaystyle x^2Division property of equality
\displaystyle \pm\sqrt{\dfrac{3V}{\pi y}}\displaystyle =\displaystyle xSquare root both sides

The solutions to this literal equation are x=\sqrt{\dfrac{3V}{\pi y}} and x=-\sqrt{\dfrac{3V}{\pi y}}.

Reflection

Literal equations are usually used in a specific context. When this is the case, we will need to consider whether any of the solutions are nonviable.

For example, if x represents a length in the equation then it cannot be negative and we would consider the solution x=-\sqrt{\dfrac{3V}{\pi y}} to be nonviable.

Example 2

A sculpture includes a cast iron parabola, coming out of the ground, that reaches a maximum height of 2.25\text{ m} , and has a width of 6\text{ m} .

Let the position of the start of the parabola be \left(0, 0\right). Let x be the horizontal distance and y be the height of the sculpture above the ground.

Determine a quadratic function that will model the shape of the parabolic sculpture.

Approach

The vertex lies half way between the two roots and has a height of 2.25 so has coordinates \left(3, 2.25\right). As we know the vertex we will write the function in vertex form f\left(x\right) = a\left(x - h\right)^2 + k. Using another known point we can solve for a. In this case we know that \left(0,0\right) lies on the parabola.

Solution

\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x-h\right)^2+kVertex form
\displaystyle f\left(x\right)\displaystyle =\displaystyle a\left(x-3\right)^2+2.25Substitute the vertex
\displaystyle 0\displaystyle =\displaystyle a\left(0-3\right)^2+2.25Substitute \left(0,0\right)
\displaystyle 0\displaystyle =\displaystyle 9a+2.25Evaluate the exponent
\displaystyle -2.25\displaystyle =\displaystyle 9aSubtraction property of equality
\displaystyle -\frac{2.25}{9}\displaystyle =\displaystyle aDivision property of equality
\displaystyle a\displaystyle =\displaystyle -\frac{1}{4}Simplify

The parabola can be modeled by the function f\left(x\right)=-\frac{1}{4}\left(x-3\right)^2+2.25

Reflection

As we knew the roots of the parabola, we could have also written the function in factored form, using a similar method. Starting with f\left(x\right)=a\left(x-6\right)x and the substituting in the values of the vertex we would find the equation in factored form is f\left(x\right)=-\dfrac{1}{4}\left(x-6\right)x.

Example 3

A ball is thrown upward and away from the top of a building. The height y (in meters) of the ball at time x (in seconds) is given by: f\left(x\right) = - 3 x^{2} + 12 x + 36

The graph of this relationship is shown.

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a

Identify and interprety the y-intercept of the graph.

Solution

The y-intercept is the height that the ball was thrown from. The ball was thrown from a height of 36\text{ m} . In the context of the question, this is also the height of the building.

Reflection

We can also see from the quadratic function used to model the flight of the ball, which is in standard form, that c=36 which corresponds with the y-intercept.

b

Identify and interpret the x-intercept shown.

Solution

The x-intercept is the time it takes for the ball to reach the ground. The ball reaches the ground after 6 seconds.

Reflection

As the domain is limited to be greater or equal to zero, the negative root can be discounted.

c

Determine the maximum height reached by the ball.

Approach

The maximum height occurs at the vertex of the parabola. As we are given the function in standard form we can find the x-value of the vertex by calculating -\dfrac{b}{2a}. Alternatively, we can read the value from the provided graph.

Solution

The vertex occurs on the line of symmetry which can be found by calculating x=-\dfrac{b}{2a}.

\displaystyle x\displaystyle =\displaystyle -\frac{b}{2a}State the line of symmetry
\displaystyle =\displaystyle -\frac{12}{2\left(-3\right)}Substitute values
\displaystyle =\displaystyle 2Simplify

The line of symmetry is x=2. We can now substitute this value in to y = - 3 x^{2} + 12 x + 36, to find the coordinates of the vertex.

\displaystyle f\left(x\right)\displaystyle =\displaystyle - 3 x^{2} + 12 x + 36State the function
\displaystyle f\left(2\right)\displaystyle =\displaystyle - 3 \left(2\right)^{2} + 12 \left(2\right) + 36Evaluate f\left(2\right)
\displaystyle =\displaystyle 48Simplify

The vertex occurs at \left(2, 48\right), which means the ball reaches a maximum height of 48 meters.

Outcomes

A2.N.Q.A.1.B

Use appropriate quantities in formulas, converting units as necessary.

A2.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

A2.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A2.A.CED.A.3

Rearrange formulas to isolate a quantity of interest using algebraic reasoning.*

A2.A.REI.A.1

Understand solving equations as a process of reasoning and explain the reasoning. Construct a viable argument to justify a solution method.

A2.F.IF.A.3

Understand geometric formulas as functions.*

A2.MP1

Make sense of problems and persevere in solving them.

A2.MP2

Reason abstractly and quantitatively.

A2.MP3

Construct viable arguments and critique the reasoning of others.

A2.MP4

Model with mathematics.

A2.MP5

Use appropriate tools strategically.

A2.MP6

Attend to precision.

A2.MP7

Look for and make use of structure.

A2.MP8

Look for and express regularity in repeated reasoning.

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