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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.)