Solving quadratic equations by graphing and factoring worksheet answers

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 0.5x, and the area of the triangle is half the product of 2 and 0.5x+2.

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.

0.25x2=0.5x+2

0.25x2−0.5x=2

0.25x2−0.5x−2=0

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.

The reflection of the y-intercept across the axis of symmetry (2,-2) is also on the parabola. With this information, the graph can be drawn.

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

The parabola intersects the x-axis twice. The points of intersection are (-2,0) and (4,0). Therefore, the quadratic equation has two solutions, x1=-2 and x2=4.

Note that x cannot be negative because 0.5x represents the side length of the square. Therefore, although x=-2 is a solution to the equation, only the solution x=4 makes sense in this context.

b The areas of the square and the triangle can be expressed as functions of x.

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

Now, plot the points (-3,2.25), (0,0), and (3,2.25) to graph f(x). Similarly, plot the points (-3,0.5), (0,2), and (3,3.5) to graph g(x).

The graphs of the functions intersect at (-2,1) and (4,4).

These points mean that the geometric shapes have the same area when x=-2 or x=4. However, since x=-2 makes the side length of the square negative, it should be discarded. Therefore, when x=4, both figures have the same area, which is 4 square units.