Math Problem Statement

how a quadratic equation works

Solution

A quadratic equation is a second-degree polynomial equation in a single variable xx, with the general form:

ax2+bx+c=0ax^2 + bx + c = 0

where aa, bb, and cc are constants with a0a \neq 0.

Key Concepts

  1. Standard Form: The standard form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0.

  2. Roots/Solutions: The solutions to the quadratic equation are the values of xx that satisfy the equation. These can be found using various methods, including factoring, the quadratic formula, and completing the square.

  3. Discriminant: The discriminant (Δ\Delta) of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is given by:

    Δ=b24ac\Delta = b^2 - 4ac

    The discriminant determines the nature of the roots:

    • If Δ>0\Delta > 0, the equation has two distinct real roots.
    • If Δ=0\Delta = 0, the equation has one real root (a repeated root).
    • If Δ<0\Delta < 0, the equation has two complex conjugate roots.
  4. Quadratic Formula: The quadratic formula provides the solutions to the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0:

    x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  5. Factoring: If the quadratic equation can be factored, it can be written as:

    (mx+n)(px+q)=0(mx + n)(px + q) = 0

    The solutions are found by setting each factor to zero:

    mx+n=0andpx+q=0mx + n = 0 \quad \text{and} \quad px + q = 0

  6. Completing the Square: This method involves rewriting the quadratic equation in the form (xp)2=q(x - p)^2 = q, making it easier to solve for xx.

Example

Given the quadratic equation x2+5x+6=0x^2 + 5x + 6 = 0:

  1. Factoring: (x+2)(x+3)=0(x + 2)(x + 3) = 0 Solutions: x=2x = -2 and x=3x = -3.

  2. Quadratic Formula: a=1,  b=5,  c=6a = 1, \; b = 5, \; c = 6 x=5±5241621=5±25242=5±12x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm 1}{2} Solutions: x=2x = -2 and x=3x = -3.

  3. Completing the Square: x2+5x+6=0x2+5x=6x^2 + 5x + 6 = 0 \quad \Rightarrow \quad x^2 + 5x = -6 Add (52)2(\frac{5}{2})^2 to both sides: x2+5x+(52)2=6+(52)2x^2 + 5x + \left(\frac{5}{2}\right)^2 = -6 + \left(\frac{5}{2}\right)^2 (x+52)2=14\left(x + \frac{5}{2}\right)^2 = \frac{1}{4} x+52=±12x + \frac{5}{2} = \pm \frac{1}{2} Solutions: x=2x = -2 and x=3x = -3.

Would you like further details or have any questions about how quadratic equations work?

Here are some related questions to expand your knowledge:

  1. How do you derive the quadratic formula from completing the square?
  2. Can you give an example of solving a quadratic equation with complex roots?
  3. What are the applications of quadratic equations in physics?
  4. How do you graph a quadratic function and find its vertex?
  5. What is the relationship between the coefficients of a quadratic equation and its graph?
  6. How can you determine if a quadratic equation can be factored easily?
  7. What are some examples of quadratic equations in real-life scenarios?
  8. How do you use the quadratic formula to solve equations with irrational solutions?

Tip: Practice solving quadratic equations using all three methods (factoring, quadratic formula, and completing the square) to become proficient and recognize which method is most efficient for different types of quadratic equations.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Polynomials
Discriminant

Formulas

Quadratic Formula
Discriminant

Theorems

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Suitable Grade Level

Grades 9-12