Quadratic Formula Study Notes

Quadratic Formula Study Notes

Overview of Quadratic Equations

  • Quadratic equations are typically in the form:
    • ax2+bx+c=0ax^2 + bx + c = 0
    • where a, b, and c are constants, and a ≠ 0.

Methods of Solving Quadratic Equations

  • Completing the Square: A method previously discussed for solving quadratic equations, especially those which cannot be factored. It involves rearranging the equation to isolate a perfect square.
  • Quadratic Formula: A formula derived from completing the square to solve any quadratic equation.

Deriving the Quadratic Formula

  • Start with the general quadratic equation:
    • ax2+bx+c=0ax^2 + bx + c = 0
  • Rearranging to isolate a perfect square:
    • a(x2+bax)+c=0a(x^2 + \frac{b}{a}x) + c = 0
    • Completing the square for the term (x2+bax)(x^2 + \frac{b}{a}x):
    • =a(x2+bax+(b/2a)2−(b/2a)2)+c= a(x^2 + \frac{b}{a}x + \frac{(b/2a)^2 - (b/2a)^2}) + c
    • Further rearranging leads to:
    • (x+b2a)2=b2−4ac4a2(x + \frac{b}{2a})^2 = \frac{b^2 - 4ac}{4a^2}
  • Therefore, the quadratic formula is:
    • x = rac{-b \pm \ rac{\sqrt{b^2-4ac}}{2a}

Quadratic Formula Usage

Example Problems

  1. Solve the equation: x2+2x−1=0x^2 + 2x - 1 = 0

    • Assign values: a=1,b=2,c=−1a = 1, b = 2, c = -1
    • Substitute into the quadratic formula:
      • x=−2±22−4(1)(−1)2(1)x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}
      • =−2±82= \frac{-2 \pm \sqrt{8}}{2}
      • =−2±222= \frac{-2 \pm 2\sqrt{2}}{2}
      • =−1±2= -1 \pm \sqrt{2}

  2. Solve the equation: 4x2−12x+3=04x^2 - 12x + 3 = 0

    • Assign values: a=4,b=−12,c=3a = 4, b = -12, c = 3
    • Substitute into the quadratic formula:
      • x=12±(−12)2−4(4)(3)2(4)x = \frac{12 \pm \sqrt{(-12)^2 - 4(4)(3)}}{2(4)}
      • =12±144−488= \frac{12 \pm \sqrt{144 - 48}}{8}
      • =12±968= \frac{12 \pm \sqrt{96}}{8}
      • =12±468= \frac{12 \pm 4\sqrt{6}}{8}
      • =3±62= \frac{3 \pm \sqrt{6}}{2}

  3. Solve the equation: 6x2+5x+1=06x^2 + 5x + 1 = 0

    • Assign values: a=6,b=5,c=1a = 6, b = 5, c = 1
    • Substitute into the quadratic formula:
      • x=−5±52−4(6)(1)2(6)x = \frac{-5 \pm \sqrt{5^2 - 4(6)(1)}}{2(6)}
      • =−5±25−2412= \frac{-5 \pm \sqrt{25 - 24}}{12}
      • =−5±112= \frac{-5 \pm 1}{12}
Practice Problems

Solve the following equations using the Quadratic Formula:

  • a. −3x2+4x+1=0-3x^2 + 4x + 1 = 0
  • b. 4x2−12x−9=04x^2 - 12x - 9 = 0
  • c. 2x2−6x+1=02x^2 - 6x + 1 = 0

Determining Intercepts of Quadratic Functions

  1. Intercepts of Function: f(x)=x2+20x+15f(x) = x^2 + 20x + 15
    • y-intercept: f(0)=15f(0) = 15
    • x-intercept: Solve for xx using:
      • 0=x2+20x+150 = x^2 + 20x + 15
      • x=−20±202−4(1)(15)2(1)x = \frac{-20 \pm \sqrt{20^2 - 4(1)(15)}}{2(1)}
      • =−20±3402= \frac{-20 \pm \sqrt{340}}{2}
      • =−10±85= -10 \pm \sqrt{85}
      • Approximate values: x=−0.78,−19.22x = -0.78, -19.22
  2. Intercepts of Function: f(x)=5x2+12x−5f(x) = 5x^2 + 12x - 5
    • y-intercept: f(0)=−5f(0) = -5
    • x-intercept: Solve for xx using:
      • 0=5x2+12x−50 = 5x^2 + 12x - 5
      • x=−12±122−4(5)(−5)2(5)x = \frac{-12 \pm \sqrt{12^2 - 4(5)(-5)}}{2(5)}
      • Approximate values yield: x=1.34,−3.74x = 1.34, -3.74

Recommended Methods for Solving Quadratic Equations

  • Graphing: Useful for approximate solutions, especially when exact values are not necessary.
  • Quadratic Formula: Best for obtaining exact solutions when required.
  • Factoring: Optimal when possible, as it simplifies the calculation significantly.

Conclusion

  • The Quadratic Formula is a powerful tool for solving quadratic equations efficiently and can be applied to a wide range of quadratic expressions. Understanding and practicing its application is crucial for mastery of quadratic equations.