Topics
3. Quadratic Equations
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United States of America
Grades 9-12

3.06 Linear-quadratic systems

Lesson
Practice

Introduction

In 8th grade, we were introduced to solving systems of linear equations graphically and algebraically. We revisited graphing systems in lesson  4.01 Writing and graphing linear systems  . Methods for solving systems of linear equations were presented to us in lesson  4.02 Solving systems using substitution  and lesson  4.03 Solving systems using elimination  . In this lesson, we will explore systems of equations that include quadratic functions.

Linear-quadratic systems

Exploration

Consider the following linear-quadratic system of equations:\begin{cases} y = 2x + 5 \\ y = x^{2} + 3x - 2 \end{cases}

  1. How could you use the substitution method to solve the system of equations?
  2. How could you use the elimination method to solve the system of equations?

A system of equations is a set of equations that have the same variables. We can apply the same algebraic methods for solving linear systems to non-linear systems. The goal is to create a one-variable equation so we can solve for that variable and use substitution to solve for the other.

We can go back and forth between representing systems of equations as a system, or as one equation, depending on the solution method we prefer to use. Consider the following equation:

\displaystyle 2x-3\displaystyle =\displaystyle x^2-x+4

This can be rewritten as:\begin{cases} y = 2x - 3 \\ y = x^{2} - x +4 \end{cases}

The solution to a system of equations is any ordered pair that makes all of the equations in the system true. For graphs, this will be the point(s) of intersection. Solutions can be found algebraically or graphically.

x
y
The line and parabola have no points of intersection, so the system has no solution.
x
y
The line and parabola have one point of intersection, so the system has one solution.
x
y
The line and parabola have two points of intersection, so the system has two solutions.

The solution to a system of equations in a given context is viable if the solution makes sense in the context and is non-viable if it does not make sense.

Examples

Example 1

Consider the following systems of equations:

\begin{cases} y= x^{2} - 2 x - 3 \\ y= - x + 3 \end{cases}

a

Graph the equations on the same coordinate plane.

Worked Solution
Create a strategy

The solution(s) to a system of equations can be represented graphically as their point(s) of intersection. We can use technology to graph the two equations, or, if drawing them by hand, it will be useful to first fill out a table of values for both equations. We can use what we know about function types to pick the best range of table values. For example, we know the vertex of the quadratic equation will be at x = \dfrac{2}{(2)(1)} so we will want to choose x-cordinates on either side of x=1.

x-3-2-101234
y=x^{2} - 2 x - 31250-3-4-305
x-3-2-101234
y=-x+36543210-1
Apply the idea
-5
-4
-3
-2
-1
1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
4
5
y
Reflect and check

Notice that from the table of values both function have the columns with the points \left(-2,5\right) and \left(3,0\right). We want these points of intersection to be visible on our graph. We also want the vertex of the parabola, \left(1,-4\right), to be visible.

We want to ensure that the x-values on our graph cover at least the interval -3 \leq x \leq 4 and the y-values on our graph cover at least the interval -5 \leq y \leq 6.

b

Identify the coordinates of the solution(s) to the system of equations.

Worked Solution
Apply the idea

The points of intersection occur at \left(-2, 5\right) and \left(3, 0\right), so these coordinate pairs will be the solutions to the system of equations.

Reflect and check

We can check our work by solving the system of equations algebraically, by equating both equations and solving for x.

\displaystyle y \displaystyle =\displaystyle x^2-2x-3First equation
\displaystyle -x+3\displaystyle =\displaystyle x^2-2x-3 Substitute y=-x+3
\displaystyle 3\displaystyle =\displaystyle x^2-x-3Add x to both sides
\displaystyle 0\displaystyle =\displaystyle x^2-x-6Subtract 3 from both sides
\displaystyle 0\displaystyle =\displaystyle \left(x+2\right)\left(x-3\right)Factor the quadratic

Using the zero product property, we can see the two solutions to this new quadratic are x=-2 and x=3.

We can now substitute these into one of the given equations to find the corresponding y-values.

Example 2

Find the solution(s) for the following linear-quadratic system of equations:\begin{cases} y = 3 x + 1 \\ y = x^{2} - 5x \end{cases}

Worked Solution
Create a strategy

We can approach this graphically or algebraically.

-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-5
5
10
15
20
25
y

But we can see from the graph that the points of intersection are not clearly identifiable. In cases like this, an algebraic approach is preferable. As both equations are already in terms of y, we can use the substitution method to solve.

Apply the idea
\displaystyle y \displaystyle =\displaystyle x^2-5xSecond equation
\displaystyle 3x+1 \displaystyle =\displaystyle x^2-5xSubstitute y=3x+1
\displaystyle 1\displaystyle =\displaystyle x^2-8xSubtract 3x from both sides
\displaystyle 0\displaystyle =\displaystyle x^2-8x-1Subtract 1 from both sides

This equation is not easily factorable, so we will use the quadratic formula to find the solutions.

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle =\displaystyle \frac{-\left(-8 \right) \pm \sqrt{\left(-8\right)^2-4\left(1 \right)\left(-1\right)}}{2\left( 1\right)}Substitute a=1, b=-8, c=-1
\displaystyle =\displaystyle \frac{8\pm \sqrt{64+4}}{2}Evaluate the square and products
\displaystyle =\displaystyle \frac{8\pm \sqrt{68}}{2}Evaluate the sum in the radicand
\displaystyle =\displaystyle \frac{8\pm 2\sqrt{17}}{2}Simplify the radical
\displaystyle =\displaystyle \frac{8}{2}\pm \frac{2\sqrt{17}}{2}Rewrite as two fractions
\displaystyle =\displaystyle 4\pm \sqrt{17}Simplify the quotients

We have found the x-coordinates of the points of intersection. We can substitute these into either equation to find the corresponding y-coordinate.

\displaystyle y\displaystyle =\displaystyle 3x+1
\displaystyle y\displaystyle =\displaystyle 3\left(4-\sqrt{17}\right)+1
\displaystyle y\displaystyle =\displaystyle 13-3\sqrt{17}

So one solution is \left(4-\sqrt{17}, 13-3\sqrt{17}\right), and using the same method, we find the other solution is \left(4+\sqrt{17}, 13+3\sqrt{17}\right).

Example 3

A base jumper jumps from the bridge of the Petronas towers, 560 ft high, immediately deploys his parachute, and then descends at a constant rate. At the same time, a ball is thrown from the observation deck of the tower, 1214 ft feet high and falls under gravity. Their height, y, after x seconds, is given by:

  • Base jumper: y=-25x+560

  • Ball: y=-16(x-1.5)^2+1250

a

Graph the height of the base jumper and ball on the same coordinate plane.

Worked Solution
Create a strategy

We can use technology to obtain a graph of the two functions, using the context and graphs to identify an appropriate domain and range for the graph.

Alternatively, we can graph by hand using key features of the graph such as:

  • The equation for the base jumper's height is a linear function, so we can identify the y-intercept, find another point on the line using the slope, and then graph the line through those two points.

  • The equation for the ball is in vertex form, so we can see the direction of opening from the coefficient, plot the vertex, then plot the y-intercept of \left(0,1214\right), and substitute another value for x to get the shape.

Apply the idea
Height over time
5
10
15
20
x \text{ (seconds)}
200
400
600
800
1000
1200
y \text{ (feet)}
Reflect and check

Consider what domain is appropriate for each graph in the context. The graphs should not extend beyond their valid domain.

b

Determine the time the base jumper and ball are at the same height.

Worked Solution
Create a strategy

Let's use technology to solve this problem. We can start by inputting the two equations for the base jumper and ball into a graphing calculator. We already have a sketch, so know we should end up with a parabola and a line.

A screenshot of the Geogebra Graphing Calculator showing the graphs of the functions y equals negative 16 quantity x minus 1.5 squared plus 1,250, and y equals negative 25 x plus 560. Speak to your teacher for more details.

We can estimate the coordinates by eye, but we should determine the coordinates more precisely using an intersection tool.

Apply the idea

From the graph, we can see that the graphs intersect at a single point in the domain appropriate to the context. To find the coordinates of this point, we can use the intersection tool. For the built-in GeoGebra graphing calculator, we need to click on the point of intersection.

A screenshot of the Geogebra Graphing Calculator showing the graphs of the functions y equals negative 16 quantity x minus 1.5 squared plus 1,250, and y equals negative 25 x plus 560. One of the points of intersection is highlighted. Speak to your teacher for more details.

The point where the graphs intersect within the restricted domain of the context is:\left(9.07, 333.26\right).

So, the ball and base jumper will be at the same height of approximately 333.26 ft at 9.07 seconds after the throw/jump.

Reflect and check

We can confirm this algebraically as well:

\displaystyle y\displaystyle =\displaystyle -25x+560First equation
\displaystyle -16\left(x-1.5\right)^2+1250\displaystyle =\displaystyle -25x+560Substitute y=-16\left(x-1.5\right)^2+1250
\displaystyle -16x^2+48x+1214\displaystyle =\displaystyle -25x+560Square the binomial and combine like terms
\displaystyle -16x^2+73x+654\displaystyle =\displaystyle 0Add 25x and subtract 560 from both sides

The solutions of this quadratic equation would give us the x-coordinates for any points of intersection between the two graphs.

The discriminant of this equation is \left(73\right)^2-4 \cdot \left(-16\right) \cdot 654=47\,185>0 which means there are two real solutions.

We could then use the quadratic formula to get the x-values of the points of intersection, ignoring negative solutions for this context.

Idea summary

A linear-quadratic system can be solved using the graphing method or substitution method. Using the graph is best when the intersection point(s) are clearly visible. Otherwise, solving the system algebraically or using an intersection tool with technology will lead to a more precise solution.

Outcomes

A.REI.C.7

Solve a simple system consisting of a linear equation and a quadratic equation in two variables algebraically and graphically.

A.REI.D.11

Explain why the x-coordinates of the points where the graphs of the equations y = f(x) and y = g(x) intersect are the solutions of the equation f(x) = g(x); find the solutions approximately.

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