Low Order Polynomials, Quadratic Equations

Quadratic Equations

A quadratic equation has degree 2, as in 3x²+5x-7 = 0. Since the latin root quad is usually associated with 4, as in quadruplets and quadrangle, you might expect a quadratic equation to have degree 4. But quad is latin for square (4 sides), hence a quadratic equation is based on a squared term, and has degree 2.

Before we solve the equation, let's find its minimum or maximum value. It always has one; minimum if a is positive and maximum if a is negative. The graph atains its extreme value when the first derivative is 0. Differentiate to get the equation 2ax+b = 0. Therefore the minimum or maximum occurs when x = -b/2a.

Now let's solve for x. In other words, where does the parabola cross the x axis?

Given any polynomial, such as a quadratic, divide through by its lead coefficient. Thus we can assume the polynomial is monic; the lead coefficient is 1.

If the next coefficient is s, replace x with y-s/n, where n is the degree. This causes the second term to drop out completely. Solve for y, then subtract s/n to get x.

When dealing with quadratics, we are done. We have y squared + a constant = 0, and that leads to y, and then to x. Let's do this for the general quadratic, to get the famous quadratic formula.

ax² + bx + c = 0

x² + (b/a)x + c/a = 0 (divide through by a)

y² + c/a + b²/(4a²) - b²/(2a²) = 0 (replace x with y-b/2a and expand)

y² = - 4ac - b² + 2b² over 4a² (common denominator)

y = ±sqrt(b²-4ac) over 2a (square root both sides)

x = -b±sqrt(b²-4ac) over 2a (back to x)

Note that this gives the right answers even if b or c is 0.

If the characteristic of the ring is 2, division by 2 is division by 0; the above procedure doesn't work.

As an exercise, build a quadratic with roots r and s. That is, (x-r)×(x-s). Apply the quadratic formula and get r and s back again.

The expression b²-4ac is sometimes called the discriminant. It is proportional to the square of the difference between the two roots. Thus the discriminant is 0 iff the two roots are identical. If we're working with real numbers, the roots are complex iff the discriminant is negative, thus taking the square root of a negative number.

Let's try a simple example, 6x² - 13x + 6 = 0.
The discriminant, b²-4ac, is 169-4×6×6, or 25. This is convenient; the square root of 25 is 5. Now the roots are 13±5 over 12, or 2/3 and 3/2. Plug these values into the quadratic; they are indeed solutions.