Quadratic Applications

Grade 9 · algebra · 100 practice problems · read aloud

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Quadratic Applications: Real-World Problem Solving

Quadratic equations aren't just abstract math! They model real-world situations involving area, projectile motion, and profit maximization. The graph of a quadratic (a parabola) helps us find maximum heights, optimal areas, and break-even points. 🚀

Step-by-Step Problem Solving Guide

  1. Understand the Problem: Identify what you're solving for (max height, time to hit ground, etc.).
  2. Set up the Equation: Translate the word problem into a quadratic equation (usually in the form ax² + bx + c = 0).
  3. Solve the Equation: Use factoring, the quadratic formula, or completing the square.
  4. Interpret the Solution: Check which answers make sense in the real-world context.

Worked Examples

Example 1: Projectile Motion

A ball is thrown upward from a height of 5 feet. Its height h (in feet) after t seconds is given by h = -16t² + 48t + 5. When does the ball hit the ground?

  1. Set h = 0: -16t² + 48t + 5 = 0
  2. Use quadratic formula: t = [-48 ± √(48² - 4(-16)(5))] / (2×-16)
  3. Simplify: t = [-48 ± √(2304 + 320)] / -32 = [-48 ± √2624] / -32
  4. Calculate: t ≈ 3.12 seconds (we discard the negative time)

Example 2: Area Problem

A rectangular garden is 10m longer than it is wide. The area is 375 m². Find the dimensions.

  1. Let width = w, length = w + 10
  2. Area equation: w(w + 10) = 375 → w² + 10w - 375 = 0
  3. Factor: (w + 25)(w - 15) = 0
  4. Solutions: w = -25 or w = 15 (only positive makes sense)
  5. Width = 15m, Length = 25m

Common Mistakes to Avoid

  • Forgetting units: Always include units (feet, seconds, meters) in your final answer.
  • Ignoring context: Discard solutions that don't make sense (negative time, negative length).
  • Misapplying formulas: Double-check the quadratic formula: x = [-b ± √(b² - 4ac)] / (2a)

Tips & Tricks

  • 💡 The vertex of the parabola gives the maximum or minimum value. For h = ax² + bx + c, the x-coordinate is -b/(2a).
  • 📐 Draw a quick sketch! Visualizing the problem helps you understand what you're solving for.
  • ⏱️ In projectile problems, the ball hits the ground when height = 0.

Practice Suggestions

  • Start with area problems (they're usually easier to visualize)
  • Practice identifying the key information in word problems
  • Create your own problems based on real situations (like throwing a ball)
  • Mix up your solving methods - try factoring, quadratic formula, and graphing

Practice problems

6 of the 100, worked through step by step — try them before opening the answer.

1 x² - 5x + 6 = 0

Hint: Look for two numbers that multiply to the constant term and add to the coefficient of the linear term.

Show the answer

Answer: x = 2, 3

  1. Identify coefficients The equation is in the form ax² + bx + c = 0. Here, a = 1, b = -5, c = 6.
  2. Factor the quadratic expression We look for two numbers that multiply to a*c = 1*6 = 6 and add to b = -5. Possible pairs for multiplication to 6: (1, 6) → sum 7 (no) (2, 3) → sum 5 (no) (-2, -3) → sum -5 (yes) So the numbers are -2 and -3.
  3. Write factored form x² - 5x + 6 = (x - 2)(x - 3) = 0.
  4. Apply the zero-product property If (x - 2)(x - 3) = 0, then: x - 2 = 0 or x - 3 = 0.
  5. Solve each equation x - 2 = 0 → x = 2 x - 3 = 0 → x = 3 Final answer: x = 2, 3

We are solving the quadratic equation: x² - 5x + 6 = 0

2 f(x) = x² - 4x + 3, f(5) = ?

Hint: Substitute the given value into the function and follow the order of operations.

Show the answer

Answer: 8

  1. Write the function: f(x) = x² - 4x + 3
  2. Substitute x = 5 into the function: f(5) = (5)² - 4(5) + 3
  3. Calculate the square: (5)² = 25
  4. Calculate the multiplication: 4(5) = 20
  5. Substitute these values: f(5) = 25 - 20 + 3
  6. Perform the subtraction: 25 - 20 = 5
  7. Perform the addition: 5 + 3 = 8

The answer is 8.

3 f(x) = x² - 4x + 3, f(2) = ?

Hint: Substitute the given value into the function and follow the order of operations.

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Answer: -1

  1. Start with the function f(x) = x² - 4x + 3
  2. Substitute x = 2 into the function: f(2) = (2)² - 4(2) + 3
  3. Calculate the exponent: (2)² = 4
  4. Calculate the multiplication: -4(2) = -8
  5. Combine all terms: 4 - 8 + 3
  6. Perform the operations from left to right: 4 - 8 = -4, then -4 + 3 = -1

The answer is -1.

4 f(x) = x² - 6x + 8, f(4) = ?

Hint: Substitute the given value into the function and follow the order of operations.

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Answer: 0

  1. Write the function: f(x) = x² - 6x + 8
  2. Substitute x = 4: f(4) = (4)² - 6(4) + 8
  3. Calculate the square: 4² = 16
  4. Calculate the multiplication: 6 × 4 = 24
  5. Substitute these values: f(4) = 16 - 24 + 8
  6. Perform the operations from left to right: 16 - 24 = -8, then -8 + 8 = 0

The answer is 0.

5 f(x) = 2x² - 8x + 6, f(3) = ?

Hint: To evaluate a function at a specific value, substitute the given number for the variable in the function's expression and follow the order of operations.

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Answer: 0

  1. Write down the function. f(x) = 2x² - 8x + 6
  2. Substitute x = 3 into the function. f(3) = 2*(3)² - 8*(3) + 6
  3. Calculate (3)². (3)² = 9
  4. Multiply 2 by 9. 2 * 9 = 18
  5. Multiply -8 by 3. -8 * 3 = -24
  6. Now substitute these values into the expression. f(3) = 18 - 24 + 6
  7. Perform the operations from left to right. First: 18 - 24 = -6 Then: -6 + 6 = 0
  8. Conclusion. Therefore, f(3) = 0.

We are given the function f(x) = 2x² - 8x + 6 and asked to find f(3). The correct answer is 0.

6 f(x) = 2x² - 8x + 3, f(3) = ?

Hint: Substitute the given x-value into the quadratic function and follow the order of operations.

Show the answer

Answer: -3

  1. Substitute x = 3 into the function: f(3) = 2(3)² - 8(3) + 3
  2. Calculate the exponent: 3² = 9
  3. Multiply: 2 × 9 = 18 and -8 × 3 = -24
  4. Combine all terms: 18 - 24 + 3
  5. Simplify: 18 - 24 = -6, then -6 + 3 = -3

The answer is -3.

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