Quadratic Formula
Solve any quadratic.
Try ToolQuadratic Word Problems
Real-world applications of quadratic equations
Learn to solve practical problems involving projectile motion, area, profit maximization, and more.
Projectile Motion
Problem 1: Ball Throw
A ball is thrown upward with velocity 64 ft/s from a height of 80 ft. When will it hit the ground? (Use \(h = -16t^2 + v_0t + h_0\))
Setup: \(-16t^2 + 64t + 80 = 0\)
Solve: Divide by -16: \(t^2 - 4t - 5 = 0\)
Factor: \((t - 5)(t + 1) = 0\)
Answer: \(t = 5\) seconds (positive root)
Problem 2: Rocket Launch
A rocket is launched with \(h(t) = -5t^2 + 30t\) meters. Find when it reaches maximum height.
Vertex: \(t = -b/(2a) = -30/(2 \times -5) = 3\) seconds
Max Height: \(h(3) = -5(9) + 30(3) = 45\) meters
Area & Geometry
Problem 3: Garden Fence
A farmer has 100 meters of fencing for a rectangular garden against a wall. What dimensions maximize the area?
Let: width = w, length = 100 - 2w
Area: \(A = w(100 - 2w) = 100w - 2w^2\)
Vertex: \(w = -b/(2a) = -100/(2 \times -2) = 25\) m
Dimensions: 25m × 50m, Area = 1250 m²
Problem 4: Rectangle Perimeter
A rectangle has perimeter 60 cm. If the area is 200 cm², find dimensions.
Let: width = w, length = 30 - w
Area: \(w(30 - w) = 200\)
Equation: \(-w^2 + 30w - 200 = 0\)
Solution: w = 10 cm or w = 20 cm
Business Applications
Problem 5: Profit Maximization
A company's profit is \(P(x) = -2x^2 + 100x - 800\) where x is items sold. Find maximum profit.
Vertex: \(x = -b/(2a) = -100/(2 \times -2) = 25\) items
Max Profit: \(P(25) = -2(625) + 2500 - 800 = 450\)
Answer: Sell 25 items for $450 profit
Problem 6: Price Demand
Demand equation: \(p = 100 - 2x\). Find revenue function and maximum.
Revenue: \(R = xp = x(100 - 2x) = 100x - 2x^2\)
Max at: \(x = 25\), Revenue = $1250
Number Problems
Problem 7: Consecutive Integers
Find two consecutive integers whose product is 72.
Let: n and n+1
Equation: \(n(n+1) = 72\)
Solve: \(n^2 + n - 72 = 0\)
Factor: \((n+9)(n-8) = 0\)
Answer: 8 and 9 (or -9 and -8)