Quadratic Equations Exam Notes

One-variable Quadratic Equations: Exam Preparation Notes

Question 1: Conditions for Real Roots

  • For the quadratic equation ((m - 1)x^{2} + 2mx + m + 3 = 0) to have two distinct real roots:
  • The discriminant (D) must be greater than zero:
    • [ D = b^2 - 4ac = (2m)^2 - 4(m - 1)(m + 3) ]
  • Expand this to find the conditions for (m):
    • [ D = 4m^2 - 4[m^2 + 3m - m - 3] = 4m^2 - 4(m^2 + 2m - 3) = 4m^2 - 4m^2 - 8m + 12 ]
    • This simplifies to (D = -8m + 12)
  • Set the discriminant greater than zero:
    • [-8m + 12 > 0]
    • Solving gives the range: (m < rac{3}{2})

Question 2: Probability of Real Roots

  • Given the equation (x^{2} + 2ax + b^{2} = 0):
  • Choose (a) from {0, 1, 2, 3}
  • Choose (b) from {0, -1, 2}
  • The condition for having real roots is (D = (2a)^{2} - 4b^{2} \geq 0)
  • Evaluate combinations:
  • Case analysis based on values of (a) and (b):
    • If (b = 0:) Any value of (a) gives real roots (p = 4/12 = 1/3)
    • If (b = -1:) Check values:
    • (a = 0:) (D = 0)
    • (a = 1:) (D = 0)
    • (a = 2:) (D > 0)
    • (a = 3:) (D > 0)
    • If (b = 2:) No values of (a) satisfy (D \geq 0)
  • Total real roots probability: ( (2 + 4 + 0) / 12 = 1/3)

Question 3: Price Reduction for Profit

  • Selling scenario: average sales 20 shirts/day at a profit of (40) each.
  • Daily profit = (20 * 40 = 800).
  • Objective: Increase daily profit to (1200):
  • Loss of (L) per shirt due to price reduction:
  • Number of shirts sold increases by 2 for every (1) unit decrease in price.
  • Let (x) be the amount reduced:
    • New selling price = (40 - x)
    • New quantity sold = (20 + 2x)
    • New profit equation:
    • ((40 - x)(20 + 2x)) must equal (1200)
    • Solve for (x) to determine the optimal price reduction.

Question 4: Function of Roots

  • For the quadratic (x^{2} - 4x + 1 = 0), roots are (x1, x2)
  • Relation of roots:
  • Sum of roots: (x1 + x2 = 4)
  • Product of roots: (x1 x2 = 1)
  • We need to find:
  • [ \frac{(x1 + x2)^{2}}{\frac{1}{x1} + \frac{1}{x2}} ]
  • [ \frac{16}{\frac{x1 + x2}{x1 x2}} = \frac{16}{4} = 4 ]

Question 5: Roots of Given Quadratic Equation

  • Given (ax^{2} + bx + c = 0) with conditions:
  • (a - b + c = 0)
  • (4a + 2b + c = 0)
  • Solve the system of equations:
  • Substitute values and find roots.
  • Use the quadratic formula:
  • (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a})
  • Determine the roots based on derived conditions.