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Understanding the Quadratic Formula |
Understanding the Quadratic Formula
Every polynomial of degree 2, given by a quadratic equation y = ax^2 + bx + c
, has either 0, 1 'double' root, or 2 real roots, which are given by the quadratic formula
x = (-b +/- sqrt(b^2 - 4ac))/2a,
and one can see from this formula, if b^2 - 4ac > 0
, then there are two distinct solutions,
if b^2 - 4ac = 0
then x = -b/2a
is a double root, and if b^2 - 4ac < 0
then there are no real roots. (If we are working over the complex numbers then there are always two roots as we can take the square root of a negative number.)
You can plug the two values into the equation ax^2 + bx + c
to verify that they really are solutions, but where does this seemingly magical formula come from and why can the roots of a quadratic formula always be computed so simply like this? Vaguely, it is because we can do some algebra to "solve for x
", and these are the solutions. Before we see this, note that if we have the two roots, say x = r
and x = s
, then we have
y = ax^2 + bx + c = a(x - r)(x - s) = a(x^2 - (r + s)x + rs) = ax^2 - a(r + s)x + ars,
so it should not be too surprising that there is a relationship between the roots r
and s
and the coefficients a, b
and c
(though the formula above is by no means clear yet).
To get the formula, note that plugging a root x
into the equation means we have
y = 0 = ax^2 + bx + c,
so finding the roots, if they exist, is the same thing as solving this equation for x
.
The most common way to derive the quadratic fomula starts with ax^2 + bx + c = 0
, divides both sides by a
and then complete's the square to get an equation where the values of x
can be written down from the resulting square, but we will use a different approach here, to avoid division which makes everything a little more bulky right away, and makes the algebra more messy throughout. We start by multiplying both sides by 4a
to try and get a square. This gives
4a^2x^2 + 4abx + 4ac = 0
Now (2ax + b)^2 = 4a^2x^2 + 4abx + b^2
, which is almost the left hand side, i.e., the left hand side of our equation is almost a square. Instead, we have
4a^2x^2 + 4abx = (2ax + b)^2 - b^2,
and so substituing this into our equation above gives
(2ax + b)^2 - b^2 + 4ac = 0.
Rearranging to solve for the square, we have
(2ax + b)^2 = b^2 - 4ac,
and we can (almost) see the quadratic formula already. Taking square roots gives
2ax + b = +/- sqrt(b^2 - 4ac),
and subtracting b
from both sides, then dividing by 2a
, gives the desired result
x = (-b +/- sqrt(b^2 - 4ac))/2a.
While this may not seem terribly special, just some algebra (and any method of solving for x
really is 'just some algebra') the quadratic formula should not be taken for granted.
For any linear polynomial y = mx + b
you can find the root x = (y - b)/m
when m
is not 0. For a cubic polynomial y = ax^3 + bx^2 + cx + d
it is more messy to solve for the roots, but a cubic formula exists, and for a quartic polynomial y = ax^4 + bx^3 + cx^2 + dx + e
it is even more complex but possible to solve for x
. However, no formula can exist for the general degree 5 of higher polynomial, so the fact that such a concise and simple formula exists for equations of degree 2 is remarkable. (The idea behind why this occurs is a surprisingly complex - but beautiful - consequence of the underlying symmetry roots of polynomials have, which follows from topics in Galois theory.)