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Wize High School Algebra II Textbook (Common Core) > Quadratic Equations & Complex Numbers

Solving Quadratic Systems

Solving Quadratic Inequalities in Two VariablesPrevious Section
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Systems of Quadratic Equations


When we have more than one quadratic equation we call it a system of quadratic equations. A solution to a system must satisfy all of the equations in a system.

How to solve

To solve a system of two quadratic equations we can
  1. set both equations equal to each other
  2. combine common terms so that it is set equal to zero
  3. solve the remaining quadratic for one variable
  4. substitute this into the system to find the second variable

Example 1
Solve the system
{y=4x2−6x−6y=3x2−9x−2\begin{cases} y = 4x^2 - 6x - 6 \\ y = 3x^2 - 9x - 2 \end{cases}

ANSWER:
First we set them equal to each other, and work to get zero on one side
4x2−6x−6=3x2−9x−2x2+3x−4=0(x+4)(x−1)=0\begin{aligned} 4x^2 - 6x - 6 &= 3x^2 - 9x -2 \\ x^2 + 3x - 4 &= 0 \\ (x + 4)(x-1) &=0 \end{aligned}
From this we get that there are two solutions for x. x=−4,1x = -4, 1.
We can lastly substitute this into an equation to find the corresponding y values.

y=4(−4)2−6(−4)−6=82\begin{aligned} y &=4(-4)^2-6(-4) - 6 \\ &=82 \end{aligned}
y=4(1)2−6(1)−6=−8\begin{aligned} y &=4(1)^2-6(1) - 6 \\ &=-8 \end{aligned}
So we have two solutions. (−4,82)(-4, 82) and (1,−8)(1, -8)

Graphically

If we graph two quadratic equations, then they might cross. The places where the graphs intersect will be the solution to the system. If the graphs do not cross, then then solutions are imaginary numbers. We can find these by solving algebraically.

Example 2
Graph the system to find the solution
{y=3x2−3xy=32x2\begin{cases} y = 3x^2 - 3x \\ y = \displaystyle\frac{3}{2}x^2 \end{cases}

ANSWER:
First we can graph each of the quadratic equations.


From this we can see that they cross in two different spots. So the solution will be the points (0,0)(0,0) and (2,6)(2,6).
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Projectile Motion



Josh and Kira are at the park tossing a baseball back and forth. After a while they want to see if one person can toss the ball, while the other person attempts to hit it with a second ball. To do this they both throw a ball at the same time, in the same forward direction, but at different heights. Josh will throw underhand while Kira throws overhand.

The height of each of their baseballs can be modeled using a quadratic equation.
Here hh is the height of the ball in meters, and tt is the time in seconds.

{h=−9.8t2+32t+3h=−9.8t2+33t+1\begin{cases} h = -9.8t^2 + 32t + 3 \\ h = -9.8t^2 + 33t + 1 \end{cases}

Using these two equations, determine if the balls ever hit each other.
If they do, when does this happen, and at what height?

ANSWER:
To solve this we'll set the two equations equal to each other at get zero to one side of the equation.

−9.8t2+32t+3=−9.8t2+33t+1−t+2=0or2−t=02=t\begin{aligned} -9.8t^2 + 32t + 3 &= -9.8t^2 + 33t + 1 \\ -t + 2 &= 0 \\ \text{or} \\ 2 - t &= 0 \\ 2 &= t \end{aligned}
This turns into a linear equation which can be solved for t directly. This gives us that t = 2.
We can then substitute this back into the system to figure out the height.

h=−9.8(2)2+32(2)+3h=27.8h = -9.8(2)^2 + 32(2) + 3 \\ h = 27.8


This means that after 2 seconds the balls will hit each other at a height of 27.8 meters.

Graphs and Matching


Match each system of equations with its graph.
A.
B.
C.


D.


{y=x2−4y=−2(x−2)2+1\begin{cases} y = x^2 - 4 \\ y = -2(x-2)^2+1 \end{cases}
{y=−x2+4y=2(x−2)2−1\begin{cases} y = -x^2 + 4 \\ y = 2(x-2)^2-1 \end{cases}
{y=x2−1y=−(x+1)2+2\begin{cases} y = x^2 -1 \\ y = -(x+1)^2+2 \end{cases}
{y=(x+2)2y=−2(x−1)2+1\begin{cases} y = (x+2)^2 \\ y = -2(x-1)^2+1 \end{cases}

Solving a Quadratic System


Solve the quadratic system of equations. You can either graph these, or solve algebraically. Type in your answer as two ordered pairs, starting with the one that has the smallest x value.

{y=x(x+6)+13y=3x2−10x+43\begin{cases} y = x(x + 6) + 13 \\ y = 3x^2 - 10x + 43 \end{cases}