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Systems of Quadratic Equations

Page history last edited by Kristen Fouss 15 years, 3 months ago

There are two options to solving systems of quadratic equations.

They are:

1. Substitution

2. Elimination

When you solve a system of quadratic equations, you are looking for ordered pairs where the two quadratics meet.

1. When using substitution, you solve for one of the variables in one quadratic equation and plug it in to the other quadratic equation.

In order to solve using substitution you first have to solve for one of the variables in one of the two equations. This picture shows how to solve for one of the variables.

Now, here is an example to solve using substitution.

Ex. $eq=x^2$ + $eq=y^2$ - 4 = 0

3x - $eq=y^2$ = 0

First, solve for X or Y in the second quadratic equation. In this equation we will solve for X first.

X = $eq=\frac{y^2}{3}$

Next, plug the equation for X into the first quadratic equation and solve for Y.

$eq=(\frac{y^2}{3})^2$ + $eq=y^2$ - 4 = 0

: $eq=\frac{y^4}{9}$ + $eq=y^2$ - 4 = 0

: 9( $eq=\frac{y^4}{9}$ + $eq=y^2$ - 4) = 0

: $eq=y^4$ + 9 $eq=y^2$ - 36 = 0

: ( $eq=y^2$ - 3) ( $eq=y^2$ +12)

: $eq=y^2$ = 3 $eq=y^2$ = -12

: y = $eq=\pm\sqrt{3}$ y = $eq=\pm\sqrt{-12}$

$eq=\emptyset$ Taking a square root of a negative number results in a nonreal number.

Now solve for X using Y.

3X - ( $eq=\pm\sqrt{3}$ )^2 = 0

: 3X - 3 = 0

: X = 1

Now just put the X's and Y's in ordered pairs and you have solved a system of quadratic equations using substitution.

(1, $eq=\pm\sqrt{3}$ )

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2. When using elimination, you first add or subtract the two equations trying to completely eliminate one of the two variables. Once you have solved for the first variable, plug it back in to one of the two quadratic equations and solve for the other variable.

Ex. - $eq=x^2$ + $eq=y^2$ + 4x - 6y + 4 = 0

$eq=x^2$ + $eq=y^2$ - 4x - 6y + 12 = 0

Solve for one of the variables. In this case, we will first solve for Y by eliminating the X's out of the equation by adding the two equations together.

2 $eq=y^2$ - 12y + 16 = 0

: 0.5(2 $eq=y^2$ - 12y + 16) = 0

: $eq=y^2$ - 6y + 8 = 0

: (y - 4) (y - 2)

: y = 4, 2

Now solve for X by plugging in our answers for Y.

y = 4 : $eq=x^2$ + $eq=(4)^2$ - 4x - 6(4) + 12 = 0 y = 2 : $eq=x^2$ + $eq=(2)^2$ - 4x - 6(2) + 12 = 0

: $eq=x^2$ + 16 - 4x - 24 + 12 = 0 : $eq=x^2$ + 4 - 4x - 12 + 12 = 0

: $eq=x^2$ - 4x + 4 : $eq=x^2$ - 4x + 4

: (x - 2) (x - 2) : (x - 2) (x - 2)

: x = 2 : x = 2

Now plug the X's and Y's into ordered pairs and you have just solved a system of quadratic equations using elimination.

(2,4) (2,2)

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If you are still somewhat confused, these links are other examples of solving a system of quadratic equations using substitution. You can use one of the following two websites to attempt to further understand how to use substitution to solve systems of quadratic equations. http://library.thinkquest.org/29292/quadratic/5systems/index.htm or http://library.thinkquest.org/29292/quadratic/5systems/index.htm

This link is a video to help further show how to solve systems of quadratic equations using substitution. Even though it isn't of quadratic equations, you are still able to see how to substitute one equation into the other.

http://video.google.com/videosearch?q=A17.4+Solving+a+System+of+Equations+by+Substitution&emb=0&aq=f#

The next link is a video to show how to solve systems of quadratic equations using elimination. Again, we were not able to find a video of solely quadratic equations, however we are still able to see how to use elimination among two equations.

http://video.google.com/videosearch?q=A17.4+Solving+a+System+of+Equations+by+Substitution&emb=0&aq=f#q=Solving%20a%20System%20of%20Equations%20by%20elimination&emb=0