Roots of a Quadratic Equation
First, let's go through a derivation of the quadratic formula (I think everyone who claims
that math is a strength of theirs should know this one):
ax 2  bx  c  0
b
a( x 2  x)  c  0
a
2
a( x 2 
b
b2
 b 
x  )c
 0 (This step is known as "completing the square.")
a
4a
 2a 
a( x 2 
b
b 2  4ac
 b 
x  ) 
a
4a
 2a 
2
b 2 b 2  4ac
) 
2a
4a
2
b
b  4ac
(x  )2 
2a
4a 2
a( x 
x
b
b 2  4ac

(Note: The plus/minus is key.)
2a
4a 2
b
b 2  4ac
)

2a
2a
 b  b 2  4ac
x
2a
x
This is the standard method (when factoring doesn't work) for determining the roots of a
quadratic equation.
Real quickly, we can see that if the discriminant ( b 2  4ac ) is positive, that the equation
has two real roots. If the discriminant is zero, then the equation has one real root of
multiplicity two. Otherwise, the equation has two complex roots that are complex
conjugates of one another. (We are assuming that a, b and c are real values.)
In general with polynomial equations with all real roots, complex roots come in pairs that
are complex conjugates of one another. (Note: the complex conjugate of a + bi is a – bi,
where a and b are real.)
Furthermore, Descartes rule of signs can be used to narrow down the total number of real
and complex roots, but for the most part, graphic calculators minimize how often these
skills are stressed in today's curriculum.
Finally, though it's beyond the scope of these notes to prove it, all polynomials of degree
n have exactly n roots total.
One question we may ask is whether or not the coefficients in a quadratic equation are
related to the roots, and if so, how. Let's just consider quadratic equations with a leading
coefficient of 1, (ie. assume a=1.)
x 2  bx  c
Now, let's assume the roots of this equation are r1 and r2. Then we can also express the
equation as follows:
( x  r1 )( x  r2 )
Clearly, these two expressions are equal, so we have:
x 2  bx  c  ( x  r1 )( x  r2 )
Now, expand the right-hand side of this equation:
x 2  bx  c  x 2  (r1  r2 ) x  r1r2
Since these two equations must always be equal, it follows that their coefficients are
equal as well. This technique is called "equating coefficients" and is quite a powerful tool
in many situations when you have two separate polynomials (or expressions in a variable)
that you know must be equal. In fact, much of solving math problems in general simply
deals with expressing the same quantity in two different ways, and then equating those
two quantities.
In this case, we find the following relationships between the roots of the equation and the
coefficients:
b  (r1  r2 ) and
c  r1r2
In certain problems, these can be quite helpful. Consider the following:
Let r and s be the roots of the quadratic equation x 2  7 x  13 . Find the quadratic
equation with a leading coefficient of one with roots 2r and 2s.
Using the relationships above, we know that r + s = -7 and rs = 13. Now, we can solve for
the two following quantities: 2r + 2s and (2r)(2s). We have:
2r + 2s = 2(r + s) = 2(-7) = -14
(2r)(2s) = 4(rs) = 4(13) = 52
It follows that the desired quadratic equation is x 2  14 x  52 .
Now, let's take a look at the cubic equation with a leading coefficient of one and roots r1,
r2, and r3.
x 3  bx 2  cx  d  ( x  r1 )( x  r2 )( x  r3 )
x 3  bx 2  cx  d  x 3  (r1  r2  r3 ) x 2  (r1 r2  r1 r3  r2 r3 ) x  r1 r2 r3
So, it follows that:
b  (r1  r2  r3 ) , c  r1r2  r1r3  r2 r3 , and d  r1r2 r3
We can see some patterns persisting here. First of all, the second coefficient is still the
negative sum of the roots. A very quick analysis of the proof above shows that this will
be true for polynomials of any degree. The constant coefficient is the negative product of
the roots for a cubic. In general, this coefficient is either the positive or negative of the
product of the roots for any polynomial. The sign depends on the degree of the
polynomial – if it's even, the sum is the positive one, otherwise it's negative.
In general, for all the other coefficients, we see that they are taken together in groups of
one, then two, then three, etc. and the signs alternate as well. Furthermore, in each group,
we go through every combination of the products of that group exactly once. This is
difficult to write out in mathematical notation, but the pattern is most definitely there.
Let's take a look at a rather difficult problem that can be solved using this knowledge:
Prove that the roots of x 5  ax 4  bx 3  cx 2  dx  e  0 can not all be real if 2a2 < 5b.
Let the roots of the equation be r1, r2, r3, r4, and r5. Then we can write down the
expressions for a and b in terms of these roots:
a  (r1  r2  r3  r4  r5 ) b  r1r2  r1r3  r1r4  r1r5  r2 r3  r2 r4  r2 r5  r3 r4  r3 r5  r4 r5
Given that 2a2 < 5b, it follows that 2a2 – 5b < 0. Now, let's examine the quantity on the
left of this inequality:
4
5
2a 2  5b  2((r1  r2  r3  r4  r5 )) 2  5(  ri r j )
i 1 j i 1
4
5
4
5
 2r1  2r2  2r3  2r4  2r5  4(  ri r j )  5(  ri r j )
2
2
2
2
2
i 1 j i 1
4
5
 2r1  2r2  2r3  2r4  2r5  (  ri r j )
2
2
2
2
2
i 1 j i 1

4
5
1
(ri  r j ) 2


2 i 1 j i 1
i 1 j i 1
This last step is very difficult to see. If you do the algebra, expanding out the last
expression, you will indeed see that it gives you the expression on the line before it.
Basically, each of the terms of the form rirj is half of the middle part of the square of the
form (ri – rj)2. When we cycle through each of these ten pairs, we see that we use each
individual term of the form ri four times, (instead of two). So we can alleviate the whole
issue by simply factoring out a one-half from the whole expression to ensure equality
with the previous line.
The reason this form is key is that if all the roots are real, then this expression is the sum
of the squares of real numbers, which we know must be non-negative. BUT, we are given
that this expression is in fact negative. If this is the case, it follows that not all of the roots
are real – at least one of them (and in reality at least two of them) must NOT be real.
This question was quite difficult. It was taken from the 1983 United States of America
Mathematical Olympiad (problem 2).
Here's an easier question that utilizes this information:
The complex number z = 1 + 2i is a solution to the equation z 3  3z 2  7 z  5  0 . Find
the other two solutions.
Since the coefficients are all real, it follows that a second root is 1 – 2i. Let the third root
be r. Using knowledge about the sum of the roots we have:
(1 + 2i) + (1 – 2i) + r = 3
r = 1.
So, the other two roots are 1 – 2i and 1.
This problem is taken from the 1997 IB HL Math Paper 1 (problem 10).