underground
mathematics
Discriminating
Problem
Below are several statements about the quadratic equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏 and
𝑐 are allowed to be any real numbers except that 𝑎 is not 0.
For each statement, decide whether MUST, MAY or CAN’T is the correct word to use in the
statement.

To say that something MUST be the case, we need it to be true in all cases; we will
need to give a convincing explanation (a proof) of why this must be always true.
To show that something CAN’T be the case, we likewise need to give a convincing
explanation (a proof) of why.
To show that something MAY be the case, we need to give an example when it is true
and an example when it is false. If you want a harder challenge, can you determine
exactly when it is and when it is not true?
(1) If 𝑎 < 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have real roots.
(2) If 𝑏2 − 4𝑎𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have one repeated real
root.
(3) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then 𝑎𝑥2 + 𝑏𝑥 − 𝑐 = 0 MUST / MAY / CAN’T
have two distinct real roots.
2
(4) If 𝑏𝑎 < 4𝑐, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have two distinct real roots.
(5) If 𝑏 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have one repeated real root.
(6) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have three real roots.
(7) If 𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have real roots.
(8) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have the same number of real
roots as 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0.
(9) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has two distinct real roots, then we MUST / MAY / CAN’T have
2
𝑎𝑐 < 𝑏4 .
(10) If 𝑐 > 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have two distinct real roots.
(11) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST / MAY / CAN’T have the same number of real
roots as 𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0.
(12) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then −𝑎𝑥2 − 𝑏𝑥 − 𝑐 = 0 MUST / MAY / CAN’T
have two distinct real roots.
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underground
mathematics
Discriminating
Suggestion

Below are several statements about the quadratic equation
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0,
where 𝑎, 𝑏 and 𝑐 are allowed to be any real numbers except that 𝑎 is not 0.
For each statement, decide whether MUST, MAY or CAN’T is the correct word to use in
the statement.



Do you have any favourite examples of quadratics with different numbers of real
roots? Having some examples to test is a good way to begin on a problem like this.
How can we use the coefficients to determine the number of real roots?
How can thinking about the graph of the quadratic 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 help us to think
about some of these statements?
As well as trying to visualise this, it may be helpful to use the interactive graph.
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underground
mathematics
Discriminating
Solution

Below are several statements about the quadratic equation
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0,
where 𝑎, 𝑏 and 𝑐 are allowed to be any real numbers except that 𝑎 is not 0.
For each statement, decide whether MUST, MAY or CAN’T is the correct word to use in
the statement.

To say that something MUST be the case, we need it to be true in all cases; we will
need to give a convincing explanation (a proof) of why this must be always true.
To show that something CAN’T be the case, we likewise need to give a convincing
explanation (a proof) of why.
To show that something MAY be the case, we need to give an example when it is true
and an example when it is false. If you want a harder challenge, can you determine
exactly when it is and when it is not true?
You may like to explore some of the scenarios on the cards using graphing software or the
interactive graph in this resource.

(1) If 𝑎 < 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have real roots.
Knowing that 𝑎 < 0 tells us that this is a “vertex up” parabola. For example, it could look like
one of these two graphs:
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The graph of the left is a sketch of 𝑦 = −𝑥2 + 1, while the one on the right is 𝑦 = −𝑥2 − 1.
This shows that if 𝑎 < 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have real roots.

(2) If 𝑏2 − 4𝑎𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have one repeated real root.
From the quadratic formula, we know that the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 (with 𝑎 ≠ 0) has
roots
𝑥=
−𝑏 ± √𝑏2 − 4𝑎𝑐
.
2𝑎
In this case, we are told that the discriminant 𝑏2 − 4𝑎𝑐 is 0, so the only real root of the
equation is 𝑥 = −𝑏
.
2𝑎
Therefore, if 𝑏2 − 4𝑎𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST have one repeated real root.

(3) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then 𝑎𝑥2 + 𝑏𝑥 − 𝑐 = 0
MUST / MAY / CAN’T have two distinct real roots.
If the equation has no real roots, then the graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 (which intersects the
𝑦-axis at 𝑐) does not cut the 𝑥-axis, and its vertex is on the same side of the 𝑥-axis as 𝑐, as
in the following sketches, the left one showing the case of 𝑎 > 0 and the right one showing
𝑎 < 0:
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The graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 − 𝑐 is obtained by translating the first graph in the 𝑦-direction until
it intersects the 𝑦-axis at −𝑐. Since the direction of the parabola does not change (it remains
“vertex up” or “vertex down”), it must now intersect the 𝑥-axis in two distinct points.
Therefore, if 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then 𝑎𝑥2 + 𝑏𝑥 − 𝑐 = 0 MUST have two
distinct real roots.
Alternatively, we can argue algebraically. Since the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real
roots, we can say something about the discriminant: we know that 𝑏2 − 4𝑎𝑐 < 0.
Since 𝑏2 is the square of a real number, it is always non-negative (that is, 𝑏2 ≥ 0), so 𝑏2 < 4𝑎𝑐
tells us that 4𝑎𝑐 is positive.
The equation 𝑎𝑥2 + 𝑏𝑥 − 𝑐 = 0 has discriminant 𝑏2 + 4𝑎𝑐, and now this must be positive.
So 𝑎𝑥2 + 𝑏𝑥 − 𝑐 = 0 has two distinct real roots.

2
(4) If 𝑏𝑎 < 4𝑐, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have two distinct real roots.
The tempting thing to do is to multiply both sides of the inequality by 𝑎, to obtain 𝑏2 < 4𝑎𝑐.
This would give 𝑏2 − 4𝑎𝑐 < 0, so that the equation couldn’t have two distinct real roots – it
would have no real roots.
Unfortunately, we don’t know whether 𝑎 is positive or negative, so when we multiply both
sides of the inequality by 𝑎, we may be changing the direction of the inequality.
If 𝑎 is positive, then 𝑏2 < 4𝑎𝑐 and the equation has no real roots.
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However, if 𝑎 is negative, then multiplying the given inequality by 𝑎 will yield 𝑏2 > 4𝑎𝑐, so
𝑏2 − 4𝑎𝑐 > 0 and the equation has two distinct real roots.
2
So if 𝑏𝑎 < 4𝑐, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have two distinct real roots, depending upon the
sign of 𝑎.

(5) If 𝑏 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have one repeated real root.
If we start by taking 𝑎 = 1, then we have the family of quadratics 𝑦 = 𝑥2 + 𝑐, which look like
this:
We can clearly see that we may have two real roots, one repeated real root (when 𝑐 = 0) or
no real roots.
So if 𝑏 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have one repeated real root.
Alternatively, we could consider this from an algebraic perspective.
The discriminant is 𝑏2 − 4𝑎𝑐 = −4𝑎𝑐, since 𝑏 = 0. For the equation to have one repeated real
root, we require the discriminant to be zero. This will be true if and only if 𝑐 = 0 (as 𝑎 ≠ 0),
so 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have one repeated real root.

(6) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have three real roots.
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The equation CAN’T have three real roots, as a quadratic equation can have at most two
roots. (We can see this from the quadratic formula or from thinking about the shape of the
graph.)

(7) If 𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have real roots.
If 𝑐 = 0, then the equation is of the form 𝑎𝑥2 + 𝑏𝑥 = 𝑥(𝑎𝑥 + 𝑏) = 0, and so 𝑥 = 0 is a real
root.
In fact 𝑥 = − 𝑎𝑏 is also a real root, so if 𝑏 ≠ 0 then the equation has two distinct real roots,
while if 𝑏 = 0, it has one repeated real root.
Alternatively, we could calculate the discriminant, 𝑏2 − 4𝑎𝑐 = 𝑏2 (since 𝑐 = 0). Since 𝑏2 is a
square, it is never negative, so the equation always has either two distinct or one repeated
real root.
So if 𝑐 = 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MUST have real roots.

(8) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have the same number of real roots as 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0.
Changing the +𝑏𝑥 term to −𝑏𝑥 in the equation 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 has the same effect as
replacing 𝑥 by −𝑥 (since 𝑎(−𝑥)2 = 𝑎𝑥2 ).
So the effect on the graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 is to reflect it in the 𝑦-axis, as in the following
sketches:
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So we see that 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 and 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0 MUST have the same number of real
roots. We can even see more than this: the roots of 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0 are the negatives of
the roots of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0.
Alternatively, we could think in terms of the discriminant.
The equations 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 and 𝑎𝑥2 − 𝑏𝑥 + 𝑐 = 0 both have discriminant 𝑏2 − 4𝑎𝑐, so
they must have the same number of real roots (whatever that number is).

(9) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has two distinct real roots, then we
2
MUST / MAY / CAN’T have 𝑎𝑐 < 𝑏4 .
If the equation has two distinct real roots, then its discriminant must be positive, that is,
𝑏2 − 4𝑎𝑐 > 0.
Now we can rearrange that to obtain 4𝑎𝑐 < 𝑏2 . We can now divide this inequality by 4
2
(which is positive) to obtain 𝑎𝑐 < 𝑏4 .
2
So if 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has two distinct real roots, then we MUST have 𝑎𝑐 < 𝑏4 .

(10) If 𝑐 > 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have two distinct real roots.
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The value of 𝑐 tells us where the quadratic crosses the 𝑦-axis. Here are some sketches of
graphs with a positive 𝑦-intercept.
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So we can see from this that if 𝑐 > 0, then 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have two distinct real
roots.
Explicit examples of quadratic equations would include 𝑥2 − 3𝑥 + 2 = (𝑥 − 1)(𝑥 − 2) = 0,
which has two distinct real roots, and 𝑥2 + 1 = 0, which has no real roots.
Alternatively, if we consider the discriminant 𝑏2 − 4𝑎𝑐, if we take 𝑎 > 0, then 4𝑎𝑐 is positive.
So if we choose 𝑏 with 𝑏2 > 4𝑎𝑐, we will have 𝑏2 − 4𝑎𝑐 > 0 (giving two distinct real roots),
while if we choose 𝑏 with 0 ≤ 𝑏2 < 4𝑎𝑐, then 𝑏2 − 4𝑎𝑐 < 0 and there are no real roots.

(11) The equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
MUST / MAY / CAN’T have the same number of real roots as 𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0.
There is no obvious way of relating the graph of 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 to that of 𝑦 = 𝑐𝑥2 + 𝑏𝑥 + 𝑎,
though it would be fine to draw them both and compare them.
We consider instead the discriminants of the two equations. The discriminant of
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 is 𝑏2 − 4𝑎𝑐, while that of 𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0 is 𝑏2 − 4𝑐𝑎. So the discriminants
are equal, and so the equations have the same number of roots.
However, we made an assumption here which may not be valid. The equation
𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 is certainly a quadratic, as 𝑎 is non-zero. However, 𝑐𝑥2 + 𝑏𝑥 + 𝑐 = 0 is only
a quadratic if 𝑐 is non-zero. But what if 𝑐 = 0? In this case, 𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0 is just the linear
equation 𝑏𝑥 + 𝑎 = 0 (assuming 𝑏 is non-zero). So if 𝑐 = 0, the equation 𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0 has
one real root, which will not be the same number of real roots as 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0.
As explicit examples, the equation 𝑥2 − 3𝑥 + 2 = (𝑥 − 1)(𝑥 − 2) = 0 has two distinct real roots.
The equation obtained by reversing the coefficients is 2𝑥2 − 3𝑥 + 1 = 0, and this factorises as
2𝑥2 − 3𝑥 + 1 = (2𝑥 − 1)(𝑥 − 1) = 0 so has roots 1 and 12 —it too has two distinct real roots.
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On the other hand, the equation 𝑥2 − 𝑥 = 0 has two distinct real roots (since
𝑥2 − 𝑥 = 𝑥(𝑥 − 1) = 0, which has roots 𝑥 = 0 and 𝑥 = 1). The equation obtained by reversing
the coefficients is 0𝑥2 − 𝑥 + 1 = 0, that is, −𝑥 + 1 = 0, and this has just one real root
(𝑥 = 1).
Therefore, the equation 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 MAY have the same number of real roots as
𝑐𝑥2 + 𝑏𝑥 + 𝑎 = 0.
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
(12) If 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then −𝑎𝑥2 − 𝑏𝑥 − 𝑐 = 0
MUST / MAY / CAN’T have two distinct real roots.
One way to think about this is that the graph of 𝑦 = −𝑎𝑥2 − 𝑏𝑥 − 𝑐 is the reflection of the
graph 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 in the 𝑥-axis. They therefore have the same number of roots as each
other.
These two graphs show examples of this. On the left, 𝑦 = (𝑥 − 1)(𝑥 − 2) has two real roots;
on the right, 𝑦 = 𝑥2 + 1 has none. It is clear that their reflections have the same number of
real roots.
Algebraically, if we multiply −𝑎𝑥2 − 𝑏𝑥 − 𝑐 = 0 by −1, we obtain 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0, which must
therefore have the same roots as the original equation. In some sense, they are essentially
the same equation.
Therefore, if 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 has no real roots, then −𝑎𝑥2 − 𝑏𝑥 − 𝑐 = 0 CAN’T have two
distinct real roots.
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