Tasks 02-03 - Quadratic & Biquadratic Equations

Section 02: Equations & Problem-Solving Strategies

Instructions

Complete these problems to master solving quadratic and biquadratic equations. Practice choosing the appropriate method (factoring, quadratic formula, or completing the square) based on the equation’s structure.

Problem 1: Converting to Zero Form (x)

Convert each equation to the form “expression = 0” and then solve:

  1. \(5x + 15 = 30\)

  2. \(3x - 7 = 2x + 8\)

  3. \(4(x - 2) = 3x + 5\)

  4. \(\frac{2x + 6}{3} = x - 1\)

  5. \(0.5x + 2.5 = 1.5x - 3.5\)

  1. \(5x + 15 = 30\)
    • Subtract 30: \(5x + 15 - 30 = 0\)
    • Simplify: \(5x - 15 = 0\)
    • Solve: \(5x = 15\)
    • \(x = 3\)
    • Check: \(5(3) + 15 = 30\) ✓
  2. \(3x - 7 = 2x + 8\)
    • Subtract 2x: \(3x - 2x - 7 = 8\)
    • Subtract 8: \(x - 15 = 0\)
    • Solve: \(x = 15\)
    • Check: \(3(15) - 7 = 38\) and \(2(15) + 8 = 38\) ✓
  3. \(4(x - 2) = 3x + 5\)
    • Expand: \(4x - 8 = 3x + 5\)
    • Move all terms: \(4x - 8 - 3x - 5 = 0\)
    • Simplify: \(x - 13 = 0\)
    • Solve: \(x = 13\)
    • Check: \(4(13 - 2) = 44\) and \(3(13) + 5 = 44\) ✓
  4. \(\frac{2x + 6}{3} = x - 1\)
    • Multiply by 3: \(2x + 6 = 3x - 3\)
    • Move all terms: \(2x + 6 - 3x + 3 = 0\)
    • Simplify: \(-x + 9 = 0\)
    • Solve: \(x = 9\)
    • Check: \(\frac{2(9) + 6}{3} = 8\) and \(9 - 1 = 8\) ✓
  5. \(0.5x + 2.5 = 1.5x - 3.5\)
    • Move all terms: \(0.5x + 2.5 - 1.5x + 3.5 = 0\)
    • Simplify: \(-x + 6 = 0\)
    • Solve: \(x = 6\)
    • Check: \(0.5(6) + 2.5 = 5.5\) and \(1.5(6) - 3.5 = 5.5\) ✓

Problem 2: Zero Product Property (x)

Use the zero product property to solve each equation:

  1. \((x - 5)(x + 3) = 0\)

  2. \((2x - 8)(3x + 6) = 0\)

  3. \(x(x - 7) = 0\)

  4. \((4x + 12)(x - 1)(x + 2) = 0\)

  5. \(3x(2x - 10) = 0\)

  1. \((x - 5)(x + 3) = 0\)
    • Either \(x - 5 = 0\) or \(x + 3 = 0\)
    • Solutions: \(x = 5\) or \(x = -3\)
  2. \((2x - 8)(3x + 6) = 0\)
    • Either \(2x - 8 = 0\) or \(3x + 6 = 0\)
    • From first: \(2x = 8\), so \(x = 4\)
    • From second: \(3x = -6\), so \(x = -2\)
    • Solutions: \(x = 4\) or \(x = -2\)
  3. \(x(x - 7) = 0\)
    • Either \(x = 0\) or \(x - 7 = 0\)
    • Solutions: \(x = 0\) or \(x = 7\)
    • Note: Don’t forget that x = 0 is a solution!
  4. \((4x + 12)(x - 1)(x + 2) = 0\)
    • Any factor can equal zero:
    • \(4x + 12 = 0\) gives \(x = -3\)
    • \(x - 1 = 0\) gives \(x = 1\)
    • \(x + 2 = 0\) gives \(x = -2\)
    • Solutions: \(x = -3, -2,\) or \(1\)
  5. \(3x(2x - 10) = 0\)
    • Either \(3x = 0\) or \(2x - 10 = 0\)
    • From first: \(x = 0\)
    • From second: \(2x = 10\), so \(x = 5\)
    • Solutions: \(x = 0\) or \(x = 5\)

Problem 3: Discriminant Analysis (x)

For each quadratic equation, calculate the discriminant and predict the number of solutions. Then solve using the quadratic formula.

  1. \(3x^2 - 7x + 2 = 0\)

  2. \(x^2 + 4x + 4 = 0\)

  3. \(2x^2 - 3x + 5 = 0\)

  4. \(-x^2 + 6x - 9 = 0\)

  1. \(3x^2 - 7x + 2 = 0\)
    • Discriminant: \(\Delta = (-7)^2 - 4(3)(2) = 49 - 24 = 25 > 0\)
    • Prediction: Two distinct rational solutions
    • Solutions: \(x = \frac{7 \pm \sqrt{25}}{6} = \frac{7 \pm 5}{6}\)
    • \(x_1 = 2\), \(x_2 = \frac{1}{3}\)
    • Check: \(3(2)^2 - 7(2) + 2 = 12 - 14 + 2 = 0\) ✓
  2. \(x^2 + 4x + 4 = 0\)
    • Discriminant: \(\Delta = 16 - 16 = 0\)
    • Prediction: One repeated real solution
    • Solution: \(x = \frac{-4 \pm 0}{2} = -2\)
    • Note: This is \((x + 2)^2 = 0\)
  3. \(2x^2 - 3x + 5 = 0\)
    • Discriminant: \(\Delta = 9 - 40 = -31 < 0\)
    • Prediction: No real solutions
  4. \(-x^2 + 6x - 9 = 0\)
    • Discriminant: \(\Delta = 36 - 36 = 0\)
    • Prediction: One repeated real solution
    • Solution: \(x = \frac{-6}{-2} = 3\)
    • Note: This is \(-(x - 3)^2 = 0\)

Problem 4: Method Selection Practice (x)

Solve each equation using the indicated method:

  1. Factor: \(x^2 - 11x + 30 = 0\)

  2. Complete the square: \(x^2 + 8x + 3 = 0\)

  3. Quadratic formula: \(3x^2 + 5x - 7 = 0\)

  4. Your choice: \(4x^2 - 12x + 9 = 0\)

  1. Factoring:
    • Find two numbers that multiply to 30 and add to -11
    • Those are -5 and -6
    • \((x - 5)(x - 6) = 0\)
    • Solutions: \(x = 5\) or \(x = 6\)
  2. Completing the square:
    • \(x^2 + 8x + 3 = 0\)
    • \(x^2 + 8x = -3\)
    • \(x^2 + 8x + 16 = -3 + 16\)
    • \((x + 4)^2 = 13\)
    • \(x + 4 = \pm\sqrt{13}\)
    • \(x = -4 \pm \sqrt{13}\)
  3. Quadratic formula:
    • \(a = 3, b = 5, c = -7\)
    • \(x = \frac{-5 \pm \sqrt{25 + 84}}{6} = \frac{-5 \pm \sqrt{109}}{6}\)
    • \(x_1 \approx 0.907\), \(x_2 \approx -2.574\)
  4. Best method: Recognize perfect square
    • \(4x^2 - 12x + 9 = (2x - 3)^2 = 0\)
    • Solution: \(x = \frac{3}{2}\) (repeated)

Problem 5: Biquadratic Equations (xx)

Solve each biquadratic equation:

  1. \(x^4 - 10x^2 + 9 = 0\)

  2. \(x^4 + 3x^2 - 28 = 0\)

  3. \(2x^4 - 9x^2 + 4 = 0\)

  1. \(x^4 - 10x^2 + 9 = 0\)
    • Let u = x²: u² - 10u + 9 = 0
    • Factor: (u - 1)(u - 9) = 0
    • u = 1 or u = 9
    • If u = 1: x² = 1, so x = ±1
    • If u = 9: x² = 9, so x = ±3
    • Solutions: x = -3, -1, 1, 3
  2. \(x^4 + 3x^2 - 28 = 0\)
    • Let u = x²: u² + 3u - 28 = 0
    • Factor: (u + 7)(u - 4) = 0
    • u = -7 or u = 4
    • If u = -7: x² = -7 (no real solutions)
    • If u = 4: x² = 4, so x = ±2
    • Real solutions: x = -2, 2
  3. \(2x^4 - 9x^2 + 4 = 0\)
    • Let u = x²: 2u² - 9u + 4 = 0
    • Using quadratic formula: u = (9 ± √(81-32))/4 = (9 ± 7)/4
    • u = 4 or u = 1/2
    • If u = 4: x = ±2
    • If u = 1/2: x = ±1/√2 = ±√2/2
    • Solutions: x = -2, -√2/2, √2/2, 2

Problem 6: Break-Even Analysis (xx)

A small bakery has the following monthly financial structure:

  • Fixed costs (rent, utilities, salaries): $8,000
  • Variable cost per dozen pastries: $4
  • Selling price per dozen: $12
  1. Write the profit equation P in terms of x dozens sold
  2. Convert the break-even condition (P = 0) to standard zero form
  3. Solve for the break-even quantity
  4. If they currently sell 800 dozen, what is their profit?
  5. How many dozen must they sell for a profit of $4,000?
  1. Profit equation:
    • Revenue = \(12x\)
    • Total Cost = \(8,000 + 4x\)
    • Profit: \(P = 12x - (8,000 + 4x) = 8x - 8,000\)
  2. Break-even in zero form:
    • Set P = 0: \(8x - 8,000 = 0\)
    • This is already in zero form
  3. Break-even quantity:
    • \(8x - 8,000 = 0\)
    • \(8x = 8,000\)
    • \(x = 1,000\) dozen
    • The bakery must sell 1,000 dozen pastries to break even
  4. Profit at 800 dozen:
    • $P = 8(800) - 8,000 = 6,400 - 8,000 = -\(1,600\)
    • They have a loss of $1,600
  5. For profit of $4,000:
    • Set P = 4,000: \(8x - 8,000 = 4,000\)
    • Convert to zero form: \(8x - 8,000 - 4,000 = 0\)
    • \(8x - 12,000 = 0\)
    • \(8x = 12,000\)
    • \(x = 1,500\) dozen

Problem 7: Complex Zero Form Applications (xxx)

A manufacturing company produces two products, A and B, with the following relationships:

  • Combined production: \(x + y = 100\) units
  • Revenue difference: The revenue from A exceeds revenue from B by $500
  • Product A sells for $30 per unit, Product B sells for $25 per unit
  1. Express y in terms of x using the production constraint
  2. Write the revenue difference equation
  3. Substitute and convert to zero form
  4. Solve for the production quantities
  5. What would happen if the revenue difference requirement changed to $1,000?
  1. Express y in terms of x:
    • From \(x + y = 100\)
    • \(y = 100 - x\)
  2. Revenue difference equation:
    • Revenue from A: \(30x\)
    • Revenue from B: \(25y\)
    • Difference: \(30x - 25y = 500\)
  3. Substitute and convert:
    • \(30x - 25(100 - x) = 500\)
    • \(30x - 2,500 + 25x = 500\)
    • \(55x - 2,500 - 500 = 0\)
    • \(55x - 3,000 = 0\)
  4. Solve for quantities:
    • \(55x - 3,000 = 0\)
    • \(55x = 3,000\)
    • \(x = \frac{3,000}{55} = \frac{600}{11} ≈ 54.55\) units of A
    • \(y = 100 - 54.55 ≈ 45.45\) units of B
    Since we need whole units:
    • A: 55 units, B: 45 units
    • Check: $30(55) - 25(45) = 1,650 - 1,125 = \(525\) (close to $500)
  5. With $1,000 difference:
    • New equation: \(30x - 25(100 - x) = 1,000\)
    • \(55x - 2,500 = 1,000\)
    • \(55x - 3,500 = 0\)
    • \(x = \frac{3,500}{55} = \frac{700}{11} ≈ 63.64\) units of A
    • \(y ≈ 36.36\) units of B
    • Or: A: 64 units, B: 36 units

Problem 8: Work Rate Problems (xxx)

Two painters can paint a house together in 12 hours. Painter A works twice as fast as Painter B.

  1. Let x = hours for Painter B to paint alone. Express Painter A’s time in terms of x
  2. Write the combined work rate equation
  3. Solve for x
  4. How long would each painter take working alone?
  5. If they work together for 8 hours, what fraction of the house remains unpainted?
  1. Express Painter A’s time:
    • If B takes x hours alone
    • A works twice as fast, so A takes x/2 hours alone
  2. Combined work rate equation:
    • B’s rate: 1/x houses per hour
    • A’s rate: 1/(x/2) = 2/x houses per hour
    • Combined rate: 1/x + 2/x = 3/x houses per hour
    • Together they complete 1 house in 12 hours
    • So: (3/x) × 12 = 1
  3. Solve for x:
    • 36/x = 1
    • 36/x - 1 = 0
    • (36 - x)/x = 0
    • 36 - x = 0 (since x ≠ 0)
    • x = 36 hours
  4. Individual times:
    • Painter B: 36 hours alone
    • Painter A: 18 hours alone
    • Check: 1/36 + 1/18 = 1/36 + 2/36 = 3/36 = 1/12 ✓
  5. After 8 hours together:
    • Combined rate: 1/12 houses per hour
    • Work completed: 8 × (1/12) = 8/12 = 2/3
    • Remaining: 1 - 2/3 = 1/3 of the house