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SOLVING BINOMIAL EQUATIONS
By Raymond White
Now that you know how to solve linear equations (for example 6 + 2x = 10) it is time to
learn something more difficult. What you are going to learn here is how to solve binomial
equations. A binomial equation is an equation that has an x2 in it; for example: x2 + 7x + 10 = 0.
You will learn three things: [1] How to factor, [2] How to complete the square, and [3]
How to use the quadratic formula. Note that there are two kinds of binomial equations: those you
can factor and those you can’t. It’s like the whole numbers, some you can factor, some you can’t.
[1] FACTORING BINOMIAL EQUATIONS
Consider the binomial x2 + 7x + 10 = 0. How would you solve it? If it can be factored,
it’s easy, and if it can’t be factored, it’s harder. So, let’s see. I write the parenthesis first:
(
)(
)=0
Then begin to fill it in, like so:
(x
)(x
)=0
Now we examine the factors of 10. What we are looking for is two numbers where their
product is equal to 10 and their sum is 7. There are only two possibilities: 1 and 10, or 2 and 5.
Both pairs satisfy the first requirement: 1×10=10 and 2×5=10. But the second requirement, well,
1+10 certainly is not 7. However 5+2=7, so maybe we have our factors, and it looks like this:
(x
5)(x
2).
Now if the signs work then we have our correct factors. So let’s see:
(x + 5)(x + 2) = 0
x2+7x+10 = 0
is the same as
So we have our correct factors. Now, how does that help us solve the equation? Here’s
how: If we solve x+5=0 for x then that value solves the entire equation because 0 × anything is 0.
Likewise, if we solve x+2=0 for x, then that value also solves the entire equation. So, x = ‒5 is a
solution but also x = ‒2 is another solution. We can verify that by plugging both values in.
(‒5)2 + 7(‒5) +10 = 25‒35+10 = 0
so ‒5 is a solution. But also
(‒2)2 + 7(‒2) +10 = 4‒14+10 = 0
so ‒2 is another solution. There are two solutions.
That was a simple problem. Let’s do a harder one.
28x2 +184x + 96 = 0
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This one is harder mainly because the x2 has a coefficient which is 28. And since 28 is
not prime there are more than just two possibilities. But let’s make it easier by dividing by 4.
7x2 + 46x +24 = 0
Since 7 is prime that makes things easier. However 24 is certainly not prime so we have a
bunch of possibilities: eight to be exact, and they are:
(7x 1)(x
(7x 24)(x
24)
1)
(7x 2)(x
(7x 12)(x
12)
2)
(7x 3)(x
(7x 8)(x
8)
3)
(7x
(7x
4)(x
6)(x
6)
4)
Each of these (1×24, 2×12, 3×8, 4×6) has a product of 24 so any of them could be the
right factors, but only one of them is. How do we find it? By trying them. 7×24+1×1 is not 46,
7×1+24×1 is not 46, and so forth until we find 7×6+4×1=46. Ahah! That’s the correct factors.
(7x + 4)(x + 6) = 0
Now we can solve the equation and get our two answers:
7x + 4 = 0
7x = ‒ 4
x = ‒ 4/7
and
x+6=0
x=‒6
[2] COMPLETING THE SQUARE
So now you know how to solve a binomial equation by factoring. That’s good, except:
what if it’s not factorable? Then what? What you do is make it factorable. Or better yet, make it
square-root-able. Consider again whole numbers: 8 is factorable (it’s 2×4), but 9 is more than
factorable, it’s square-root-able (it’s 3×3). So, we are going to take an unfactorable binomial and
make it factorable and square-root-able. How do we do that? You’ll see with this example:
4x2 + 16x – 65 = 0
This is definitely not factorable. You can try to factor it, but you’ll be wasting your time.
So let’s try something else, let’s make it square-root-able. Well, not the whole thing, just the first
two terms (those terms that have the x’s) which is the part that matters.
First off all, get rid of the 65. Then second, get rid of the 4. Like so:
4x2 + 16x = 65
x2 + 4x = 65/4
Now what we want to do is add a magic number to the left side that makes the left side
square-root-able. How do we do that? Here’s how: We look at the coefficient of x (which is 4),
2
divide it by 2, and square it. So: 4 / 2 = 2, then 2 squared is 4. That’s the magic number. Add it to
the left side, and at the same time of course add it to the right side. And that gives us:
x2 + 4x + 4 = 65/4 + 4
Now the left side is square-root-able and the equation looks like this:
(x + 2)2 = 65/4 + 4
Go ahead, try it. Square x + 2 and you will get x2 + 4x + 4. Amazing, isn’t it?
Now let’s simplify the right side.
(x + 2)2 = 65/4 + 16/4 = 81/4
Now we can take the square root of the left side (that’s what square-root-able means), and
the right side —
_______
____
2 =
√ (x + 2)
√ 81/4
And we get
x + 2 = ± 9/2
And finally we get our two answers:
x = ‒2 + 9/2 = 2½
and
x = ‒2 ‒ 9/2 = ‒6½
And that is how it’s done, how you solve unfactorable binomial equations by Completing
the Square.
[3] A GENERAL FORMULA FOR ALL BINOMIAL EQUATIONS
Since we can now solve factorable and non-factorable binomial equations, can we use
that Completing the Square method to find a general formula that can solve all binomial
equations? Yes, we can.
Instead of specific numbers, can we use any numbers? Instead of an equation like
4x2 + 16x – 65 = 0
can we generalize the equation by using any numbers instead? Like this:
ax2 + bx + c = 0
Now, if we can solve that, so that x = f(a,b,c), then we will have our general formula.
So let’s do it. We will repeat all the steps of [2] but using a, b, and c for our numbers.
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First, get rid of c, then get rid of a.
ax2 + bx + c = 0
ax2 + bx = ‒c
x2 + (b/a)x = ‒(c/a)
Now we have to find our magic number that makes the left side square-root-able. And
that number is one half of (b/a) squared, or properly written, is: (b/2a)2 which is also b2/4a2.
So, let’s add that magic number to left side, but also to the right side. We get:
x2 + (b/a)x + (b/2a)2 = ‒(c/a) + (b/2a)2
Now that the left side is square-root-able, we can:
1. Factor the left side to its square-root-able form,
2. Simplify the right side,
3. Square root both sides.
(x + b/2a)2 = b2/4a2 ‒ c/a
= b2/4a2 ‒ 4ac/4a2
= (b2 – 4ac) / 4a2
Now we can take the square root of both sides.
_________
_____________
b
2
√ (x + /2a) = √ (b2 – 4ac) / 4a2
Which gives us
_______
b
x + /2a = √ b2 – 4ac / 2a
Now we isolate x
_______
x = ‒ b/2a ± √ b2 – 4ac / 2a
And, by a nice gift from nature, we already have our common denominator (2a) so we
combine to get our final formula:
_______
x = ‒b ± √ b2 – 4ac
2a
And that is the quadratic formula and how it is derived. Beautiful, isn’t it?
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