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The graph of an equation in two variables shows the set of all possible solutions. In this lesson, equations involving quadratic expressions will be solved by graphing.

### Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

## Finding Solutions of Quadratic Equations

Consider a quadratic function in standard form.
Recall the steps followed to graph a quadratic function in standard form. Use the applet to draw the graph.
Think about the definitions of an intercept and a solution of an equation. How do the intercepts of the quadratic function and the solutions of the quadratic equation relate to each other?
Discussion

A quadratic equation is a polynomial equation of second degree, meaning the highest exponent of its monomials is A quadratic equation can be written in standard form as follows.

Here, the leading coefficient is non-zero, which guarantees that the term is present. Solving a quadratic equation means finding the zeros of the related quadratic function. Therefore, quadratic equations can have at most two solutions.
This type of equation can be solved using several methods, such as graphing and factoring.
Method

## Solving a Quadratic Equation Graphically

A quadratic equation can be solved graphically by drawing the parabola that corresponds to the related quadratic function and identifying its zeros. Consider the following equation as an example.
These four steps can be followed to find the solutions of the equation.
1
Write the Equation in Standard Form
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If neither side of the equation equals rearrange it so that all terms are on the same side.
If there is more than one term with the same degree, simplify the equation by combining like terms.
The equation is in standard form now.
2
Graph the Related Function
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The side with the variables can now be seen as a function
To graph the quadratic function in standard form, its characteristics should be identified first.
Direction Vertex Axis of Symmetry intercept
upward

In addition to these points, the reflection of the intercept across the axis of symmetry is at which is also on the parabola. Now, the graph can be drawn.

3
Identify Any Points With coordinate
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Now, find all the points on the graph that have a coordinate equal to

4
Identify the coordinates of Those Points
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The coordinates of the identified points solve the equation

In the example, the coordinates of the points where the curve intercepts the axis are and Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into and evaluate the expression. Verifying the solution is done by evaluating
Evaluate right-hand side
When the value of is Therefore, is an exact solution. The solution can now be verified by following the same procedure.
Solution Substitute Evaluate
Since is also a solution. Therefore, the equation has two solutions.
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.
If the function has two zeros, the equation 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.
Pop Quiz

## Using a Graph to Find the Number of Solutions of a Quadratic Equation

For each quadratic equation, draw its related quadratic function to determine the number of solutions.

Example

## Finding the Time When a Rocket Reaches a Certain Height

Paulina models the flight of a rocket she built using a quadratic function, where is the height of the rocket in meters after seconds.

a After how many seconds does the rocket reach a height of meters?
b After how many seconds does the rocket reach a height of meters?

### Hint

a Set the quadratic function equal to Then, rewrite it in standard form to solve by graphing.
b Set the quadratic function equal to Then, rewrite it in standard form to solve by graphing.

### Solution

a To find after how many seconds the rocket is at a height of meters, will be substituted for into the given quadratic function.
This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.
Now the function related to the equation can be written.
The solutions to the equation are the intercepts of the function's graph. To graph the related quadratic function, its characteristics must be first determined.
Direction Vertex Axis of Symmetry intercept

In addition to the vertex and the intercept, the reflection of the intercept across the axis of symmetry, which is at is also on the parabola. With this information, the graph of the function can be drawn.

The parabola intersects the axis twice. The points of intersection are and
Therefore, the equation has two solutions, and
These solutions mean that the rocket is at a height of meters after and after seconds.
b Following the same steps as above, the time at which the rocket is at meters can be calculated.
This quadratic equation will now be solved by graphing. Before doing so, it should be rewritten in standard form.
Now the function related to the equation can be written.
Again, the solutions to the equation are the intercepts of the parabola. Identify the characteristics of the function to be able to draw it.
Direction Vertex Axis of Symmetry intercept

The reflection of the intercept across the axis of symmetry is at which is also on the parabola. Now, the graph can be drawn.

This time, there is only one intercept.

The point of intersection is Therefore, the equation has one solution,
This solution means that the rocket is at a height of meters after seconds.
Example

## Modeling Areas of Geometric Figures With Quadratic Equations

Tadeo wants to form a square and a right triangle whose areas are equal. The side lengths of the geometric shapes are expressed in the diagram.

a Write a quadratic equation in standard form to model the situation and solve it by graphing.
Write the solution(s) to the equation.
b Let and be the functions representing the areas of the square and the triangle, respectively. Express these functions in terms of
Determine the points of intersection of the functions by graphing. Write the point with the smaller coordinate first.

### Hint

a The area of a square equals its side length squared. The area of a right triangle is half the product of its legs.
b The function for the area of the triangle is a linear function. To draw its graph, make a table of values.

### Solution

a Using the given diagram, an expression for the area of each shape can be written.

The area of the square is the square of and the area of the triangle is half the product of and

Area of Square Area of Triangle
Since the geometric shapes have the same area, equating these expressions will produce a quadratic equation.
To write it in standard form, the Properties of Equality will be used. This quadratic equation will now be solved by graphing. To do so, the quadratic function related to the equation will be written.
Next, the graph of the related function will be drawn. Identify its characteristics.
Direction Vertex Axis of Symmetry intercept

The reflection of the intercept across the axis of symmetry is also on the parabola. With this information, the graph can be drawn.

From the graph, the intercepts of the function can be identified.

The parabola intersects the axis twice. The points of intersection are and Therefore, the quadratic equation has two solutions, and
Note that cannot be negative because represents the side length of the square. Therefore, although is a solution to the equation, only the solution makes sense in this context.
b The areas of the square and the triangle can be expressed as functions of
Area of the Square Area of the Triangle

To draw the graphs of these functions, a table of values will be made.

Now, plot the points and to graph Similarly, plot the points and to graph
The graphs of the functions intersect at and
These points mean that the geometric shapes have the same area when or However, since makes the side length of the square negative, it should be discarded. Therefore, when both figures have the same area, which is square units.
Example

## Modeling Heights of Arrows With Quadratic Equations

Tiffaniqua and Ramsha are good at archery. They want to determine whether the arrows they shoot will collide in the air or not. Tiffaniqua takes her shot from a tree and Ramsha takes her shot from the ground just below Tiffaniqua. They wrote two quadratic functions to model the heights of the arrows in meters.
In these equations, represents the horizontal distance in meters from the point where the arrow is shot.
a Write a quadratic equation in standard form that will help determine whether the arrows will collide.
b Find the answer that fits the context of the situation by solving the equation graphically.
c Suppose a coordinate plane is placed on the diagram so that Ramsha is at the origin. How far is Ramsha from the point where the arrows collide? Write the answer in exact form.

### Hint

a At the collision point of the arrows, the functions have the same height. Therefore, equate the given quadratic functions.
b The equation written in the previous part has a related function. Draw the graph of this function.
c Find the coordinates of the point where the arrows collide. Then, use the Distance Formula.

### Solution

a At the collision point of the arrows, if there is any, the functions have the same value. Therefore, to find this point, the right-hand sides of the equations can be equated.
This resulting quadratic equation needs to be written in standard form.
Simplify right-hand side
The solutions to this quadratic equation will give the horizontal distance to the collision point.
b The quadratic equation found in Part A will solved graphically. To do so, its related function will be drawn first.
To be able to graph the function, some of its characteristics should be determined.
Direction Vertex Axis of Symmetry intercept
The reflection of the intercept across the axis of symmetry is also on the parabola. Since these points are very close to each other, another point can be plotted by randomly choosing an value and calculating its corresponding value. Try
Evaluate right-hand side
Therefore, the point lies on the curve. The reflection of this point across the axis of symmetry is which also lies on the parabola. With this information, the graph can be drawn.

The intercepts of the graph can now be identified.

The parabola intersects the axis twice. From the graph only one of the intercepts is identified easily, which is The other intercept is between and However, this intercept can be ignored because represents horizontal distance and distance cannot be negative. Therefore, in this context, only makes sense.
This solution means that the horizontal distance from where Tiffaniqua and Ramsha stand to the point of collision is meters.
c Place a coordinate plane so that Ramsha is at the origin. In Part B, the horizontal distance to the point where the arrows collide was found to be meters
To find how far Ramsha is to the collision point, the Distance Formula will be used.
To do so, the coordinate of the collision point needs to be found. To accomplish this, substitute into the either of the quadratic functions, and evaluate.
Evaluate right-hand side