Professor Dave here, let’s derive the quadratic formula.
For several tutorials we have been learning how to solve quadratics.
Some are easy.
Four X squared equals sixty-four requires just a few steps.
Working from the outside in, we divide both sides by four, take the square root, and X
equals plus or minus four.
If we take the same expression but add an X term, it gets trickier.
We just saw how we can sometimes solve polynomials by factoring or by completing the square,
but there are times when neither of these strategies will be especially easy or even
possible to perform.
In general, it would be nice to have another method where we can solve by brute force rather
than having to perform some algorithm.
We need an equation where we can simply plug in the coefficients of the expression and
get the solution.
Fortunately, there is such an equation.
It’s called the quadratic formula.
But instead of simply looking at it, let’s derive the equation, as derivations always
help us appreciate the beauty of math that much more.
Let’s start with a general polynomial equation, AX squared plus BX plus C equals zero, where
A is not equal to zero.
Let’s try to complete the square, just as we learned before, but with this general form
instead of a specific example.
Remember, the first thing we would do is divide everything by A, because it’s easier if
there is no coefficient operating on the X squared term.
That means we get X squared plus B over A times X plus C over A equals zero.
Then as usual, we bring the last term to the other side of the equation.
Now remember that in order to complete the square, we take this term, cut it in half,
and then square it.
It may seem trickier with variables instead of numbers, but it’s really the same thing,
so we get B over two A, since dividing by two just puts a two in the denominator, and
then squaring gives us B squared over four A squared.
This term will be added to both sides.
Now on the left, we have a perfect square, so let’s factor it and get X plus B over
two A quantity squared, while on the right, we can multiply C over A by four A over four
A so that we end up with the same denominator for both fractions, allowing us to combine
them into one fraction.
Now we can take the square root of both sides.
For the left, that just gets rid of the exponent.
On the right, instead of one big radical, we can put one on the numerator and one on
the denominator.
The numerator can’t be simplified, so let’s leave the plus/minus there, but the denominator
is a perfect square, so we can change that into two A. Now we just subtract this term
on the left from both sides to get X alone, recognize that these have the same denominator
so we combine them, and there we have it, the famous quadratic formula.
Even though you just learned how to derive this, it is still a good idea to memorize
it, so repeat after me.
X equals the opposite of B plus or minus the square root of B squared minus four AC all
over two A. The reason this equation is useful is that we can take any quadratic, plug A,
B, and C into this formula, and it will spit out the solutions to the equation.
Let’s try one example now.
X squared plus two X minus three.
Now we should be able to see from our knowledge of factoring that this can easily be expressed
as X minus one times X plus three, meaning that the solutions are one and negative three.
But let’s plug the coefficients into the quadratic formula to make sure it works.
A is one, B is two, and C is negative three, so let’s just replace all of these variables
with the corresponding numbers.
Now we just simplify step by step.
Let’s combine everything under the radical to get sixteen, take the square root which
is four, and now we have two answers, negative two plus four over two, or negative two minus
four over two.
These give us two over two and negative six over two, which is one and negative three,
so the formula works.
There are analogous formulas for cubic and quartic functions, but they are pretty complicated,
so we can just stick to this one for quadratics.
Remember, this can be used for any quadratic, so when factoring is impossible, or you just
don’t feel like doing it, this formula will be your friend.
The arithmetic can get a little messy sometimes, and if you end up with a negative number for
the discriminant, which is the B squared minus four AC term under the radical, you will have
to conclude that there are no real solutions to the quadratic, since we can’t take the
square root of a negative number.
In such a case, you could express the solution using imaginary numbers, which we will learn
about later.
But whatever the case, if you prefer this method to actually completing the square yourself,
then memorize the formula and plug away.
Let’s check comprehension.