Master the art of solving systems that include quadratic equations! Learn substitution methods, graphical solutions, and how to handle multiple solution scenarios.
• One quadratic equation
• One linear equation
• Can have 0, 1, or 2 solutions
• Example: y = x² and y = 2x + 1
• Two quadratic equations
• Can have 0, 1, 2, 3, or 4 solutions
• More complex to solve
• Example: y = x² and y = -x² + 4
No Solutions
Graphs don't intersect
One Solution
Graphs touch at one point
Two Solutions
Graphs intersect twice
• Step 1: Solve one equation for one variable
• Step 2: Substitute into the other equation
• Step 3: Solve the resulting equation
• Step 4: Find the other variable
• Step 5: Check solutions in both equations
Solve the system:
y = x² - 2x + 1
y = 2x - 3
Step 1: Both equations are already solved for y
Set them equal: x² - 2x + 1 = 2x - 3
Step 2: Rearrange to standard form
x² - 2x + 1 - 2x + 3 = 0
x² - 4x + 4 = 0
Step 3: Solve the quadratic equation
x² - 4x + 4 = (x - 2)² = 0
x = 2 (double root)
Step 4: Find y-coordinate
y = 2(2) - 3 = 4 - 3 = 1
Solution: (2, 1)
This is a tangent point - the line touches the parabola at exactly one point.
Solve the system:
y = x² - 4
y = x + 2
Step 1: Set equations equal
x² - 4 = x + 2
Step 2: Rearrange to standard form
x² - 4 - x - 2 = 0
x² - x - 6 = 0
Step 3: Factor and solve
x² - x - 6 = (x - 3)(x + 2) = 0
x = 3 or x = -2
Step 4: Find y-coordinates
When x = 3: y = 3 + 2 = 5
When x = -2: y = -2 + 2 = 0
Solutions: (3, 5) and (-2, 0)
The line intersects the parabola at two points.
• Step 1: Graph both equations on the same coordinate plane
• Step 2: Identify intersection points
• Step 3: Read coordinates of intersection points
• Step 4: Verify solutions algebraically
System:
y = x² - 2x - 3
y = -x + 1
Parabola: y = x² - 2x - 3
Vertex: (1, -4)
X-intercepts: (-1, 0) and (3, 0)
Y-intercept: (0, -3)
Line: y = -x + 1
Slope: -1
Y-intercept: (0, 1)
X-intercept: (1, 0)
Intersection Points:
From the graph: (-1, 2) and (4, -3)
Verification:
Check (-1, 2): 2 = (-1)² - 2(-1) - 3 = 1 + 2 - 3 = 0 ✗
Check (4, -3): -3 = (4)² - 2(4) - 3 = 16 - 8 - 3 = 5 ✗
Need to solve algebraically for exact values!
Solve the system:
y = x² + 2x + 3
y = x - 1
Step 1: Set equations equal
x² + 2x + 3 = x - 1
Step 2: Rearrange to standard form
x² + 2x + 3 - x + 1 = 0
x² + x + 4 = 0
Step 3: Check discriminant
b² - 4ac = 1² - 4(1)(4) = 1 - 16 = -15
Since -15 < 0, no real solutions
Result: No solution
The line and parabola don't intersect. The parabola is always above the line.
Problem: A ball is thrown with height h(t) = -16t² + 32t + 5. A target is at height 21 feet. When will the ball hit the target?
System:
h = -16t² + 32t + 5
h = 21
Solution:
21 = -16t² + 32t + 5
0 = -16t² + 32t - 16
0 = -16(t² - 2t + 1) = -16(t - 1)²
t = 1 second
Result: The ball hits the target at exactly 1 second.
Always substitute your solutions back into both original equations to verify
Quadratic systems can have 0, 1, or 2 solutions - don't assume there's always exactly one
Be careful with signs when moving terms from one side to the other
Problem 1:
Solve: y = x² - 1 and y = 2x + 2
x² - 1 = 2x + 2
x² - 2x - 3 = 0
(x - 3)(x + 1) = 0
Solutions: (3, 8) and (-1, 0)
Problem 2:
Solve: y = x² + 4x + 4 and y = x + 6
x² + 4x + 4 = x + 6
x² + 3x - 2 = 0
Using quadratic formula: x = (-3 ± √17)/2
Two solutions: approximately (0.56, 6.56) and (-3.56, 2.44)
Problem 3:
Solve: y = x² + 2x + 5 and y = x - 1
x² + 2x + 5 = x - 1
x² + x + 6 = 0
Discriminant: 1 - 24 = -23 < 0
No real solutions