9.7 Solve Quadratic Systems

Quadratic system

A system that includes one or more equations of conies

Example 1

Solve a linear-quadratic system by graphing

Solve the system using a graphing calculator.

y² - 3x - 1 = 0 Equation 1

2x - y = 6 Equation 2

1. Solve each equation for y.

Equation 1 Equation 2

y² - 3x-l = 0 2x - y=6

y² = _3x +1__ - y = __-2x + 6__

y = _________ y = __2x - 6_

2. Graph the equations. Use the calculator's intersect feature to find the coordinates of the intersection points. The graphs of y = ___________ and y = 2x - 6 intersect at (_1.75_, _-2.5_). The graphs of and y = ___________ intersect at ( _5_, __4_).

The solutions are and (_1.75_, _-2.5_). Check the solutions by substituting the coordinates of the points into each of the original equations.

Example 2

Solve a linear-quadratic system by substitution

Solve the system using substitution.

x² + y² = 13 Equation 1

y = x + 5 Equation 2

Solution

Substitute x + 5 for y in Equation 1 and solve for x.

x^2 + y^2 = 13	Equation 1
x^2 + (__x + 5__)^2= 13	Substitute for y.
x^2 + x^2 + 10x +25 = 13	Expand the power.
_2x^2 + 10x + 12_ =0	Combine like terms.
__x^2 + 5x + 6__ = 0	Divide each side by _-2_.
_(x + 2)(x + 3)__ = 0	Factor.
X = _-2_ or x = _-3_	Zero product property.

The corresponding y-values are y = _-2 + 5__ = _3_ and y = _-3 + 5__ = _2_. The solutions are (_-2_,_2_) and ( _-3_ ,__3_).

Example 3

Solve a quadratic system by elimination

Solve the system by elimination.

5x² + y² + 48x + 99 = 0 Equation 1

x² - y² - 9 = 0 Equation 2

Solution

Add the equations to eliminate the y^2-term and obtain a quadratic equation in x.

5x^2 + y^2 + 48x + 99 = 0
x^2 - y^2- 9 = 0
6x^2 + 48x +90 = _0_	Add.
x^2 + 8x + 15 = _0_	Divide each side by _6__.
(x + 3)(x + 5) = _0_	Factor.
x = _-3_ or x = _-5_	Zero product property

When x = _-3_,y = _0_. When x = _-5__, y =__±4_.

The solutions are _(-3, 0)_, _(-5, 4)_, and__(-5, -4)__.

Example 4

Treasure Hunt The class is having a treasure hunt that begins at the school.

Clue 1: The treasure is 3 km from the school.

Clue 2: The treasure is 5 km from the post office. (The post office is 4 km west and 6 km north of the school.)

Clue 3: The treasure is 1 km from the town square. (The town square is 2 km north of the school.)

Solution

Let each unit represent 1 km. If the school is at (0, 0), then the post office is at _(-4, 6)_ and the town square is at _(0, 2)_. Write equations of circles that represent the possible locations of the treasure. The intersection of the circles is the location of the treasure.

Clue 1: _x² + y² = 9_

Clue 2: _(x + 4)² + (y - 6)² = 25_

Clue 3: _x² + (y - 2)² = 1_

Expand the equation from Clue 3, subtract Clue 1 from Clue 3, and solve for y.

_x² + y² - 4y + 4 = 1_

-(x² + y² = 9)

__-4y + 4 = -8_ y = _3_

Use y = _3_and the equation from Clue 1 to find x = _0_. The treasure is _3 km north_ of the school.