Overview

In many algebra problems, we are not asked to find the exact roots of a quadratic equation. Instead, we are asked to determine where the roots lie.

For example, a problem might ask:

• Are both roots greater than a certain number?
• Are both roots less than a certain number?
• Is exactly one root inside a given interval?
• Do both roots lie between two numbers?

These questions are called root-location problems.

They are especially common when a quadratic contains a parameter. Instead of solving for the roots directly, we write inequalities that force the roots to lie in the desired region.

Motivation

Consider a quadratic function

where $A\neq 0$.

The roots of the quadratic equation

are the points where the graph of $f(x)$ intersects the $x$-axis.

So locating the roots is really a graph question:

Where does the parabola cross the $x$-axis?

For example, if we want exactly one root between $0$ and $3$, we want the graph to cross the $x$-axis once inside that interval and once outside it.

Rather than drawing a new graph every time, we can use a few standard conditions involving:

• The discriminant
• The value of the function at endpoints
• The vertex of the parabola

Important Notation

Throughout this module, let

where $A\neq 0$.

Let the roots be

The discriminant is

The $x$-coordinate of the vertex is

The sign of $A$ tells us whether the parabola opens upwards or downwards.

• If $A > 0$, the parabola opens upwards
• If $A < 0$, the parabola opens downwards

The expression $Af(k)$ is useful because it compares the sign of the value $f(k)$ with the direction of the parabola.

Why the Discriminant Matters

A quadratic has real roots if and only if

If

then the graph does not cross the $x$-axis at all, so it makes no sense to ask whether the real roots are greater than or less than some number.

Therefore, whenever we want both roots to be located somewhere, we usually need

In some cases, this condition is automatic from the other inequalities, but it is still useful to understand why it appears.

Case 1: Both Roots Greater Than $k$

Suppose we want

Then both roots lie to the right of $k$.

The required conditions are:

and

Why These Conditions Work

First, we need real roots, so

Next, if both roots are to the right of $k$, then the point $x=k$ lies outside the root interval. At such a point, the graph has the same sign as the leading coefficient. This is captured by

Finally, the vertex lies halfway between the two roots, so if both roots are greater than $k$, the vertex must also lie to the right of $k$:

Case 2: Both Roots Less Than $k$

Suppose we want

Then both roots lie to the left of $k$.

The required conditions are:

and

This is very similar to the previous case. The only difference is that the vertex must now lie to the left of $k$.

Case 3: One Root on Each Side of $k$

Suppose we want one root less than $k$ and the other root greater than $k$:

The required condition is

Why Only One Condition?

If $A>0$, the parabola opens upwards. Between the two roots, the graph lies below the $x$-axis, so $f(k)<0$.

If $A<0$, the parabola opens downwards. Between the two roots, the graph lies above the $x$-axis, so $f(k)>0$.

In both cases, $A$ and $f(k)$ have opposite signs, so

This condition already forces the graph to cross the $x$-axis on both sides of $k$, so we do not need to separately write

Two-Point Root Location

Now suppose we have two real numbers

Many problems ask us to describe where the roots lie relative to the interval

There are three especially important cases:

• Both roots lie inside the interval
• Exactly one root lies inside the interval
• One root lies to the left of the interval and the other lies to the right

Case 4: Both Roots Between $k_1$ and $k_2$

Suppose we want

That means both roots lie inside the interval.

The required conditions are:

and

Why These Conditions Work

We need real roots, so

The endpoints $k_1$ and $k_2$ must both lie outside the root interval. Therefore the graph at both endpoints must have the same sign as the leading coefficient:

and

Finally, the vertex lies halfway between the two roots. If both roots are inside the interval, the vertex must also be inside the interval:

All four conditions are needed. If the vertex condition is missing, it is possible for both roots to lie outside the interval while still having the correct endpoint signs.

Case 5: Exactly One Root Between $k_1$ and $k_2$

Suppose we want exactly one root in the interval

The required condition is

Why This Works

If

then $f(k_1)$ and $f(k_2)$ have opposite signs.

Since a quadratic function is continuous, the graph must cross the $x$-axis somewhere between $k_1$ and $k_2$.

A quadratic has at most two real roots. If the signs at the two endpoints are opposite, the number of crossings inside the interval is odd. Therefore, there is exactly one root in the interval.

This is often the simplest root-location case.

Case 6: One Root on Each Side of the Interval

Say we want

That means the entire interval lies between the two roots.

The required conditions are:

and

Why These Conditions Work

If $A>0$, the parabola opens upwards and is negative between its two roots. As such, if both $f(k_1)$ and $f(k_2)$ are negative, both points lie between the roots.

If $A<0$, the parabola opens downwards and is positive between its two roots. As such, if both $f(k_1)$ and $f(k_2)$ are positive, both points again lie between the roots.

Both situations are captured by

and

This guarantees one root to the left of $k_1$ and the other root to the right of $k_2$.

Summary of Conditions

Let

with $A\neq 0$, discriminant

and vertex $x$-coordinate

One-Point Cases

Suppose $h=-\frac{B}{2A}$.

Both roots greater than $k$

If both roots are greater than $k$, use the conditions

and

Both roots less than $k$

If both roots are less than $k$, use the conditions

and

One root on each side of $k$

If one root is less than $k$ and the other root is greater than $k$, use the condition

This condition already guarantees that the parabola crosses the $x$-axis on both sides of $k$.

Two-Point Cases

Assume

Both roots in $(k_1,k_2)$

If both roots lie inside the interval $(k_1,k_2)$, use the conditions

and

Exactly one root in $(k_1,k_2)$

If exactly one root lies inside the interval $(k_1,k_2)$, use the condition

This means the function has opposite signs at the two endpoints.

One root left of $k_1$, one root right of $k_2$

If one root lies to the left of $k_1$ and the other lies to the right of $k_2$, use the conditions

and

This means both endpoints lie between the two roots.

Worked Examples

Example 1

Find all real values of $t$ such that both roots of

$$x^2-2tx+t+3=0$$

are greater than $1$.

Let

Here,

We want both roots to be greater than $1$, so we use Case 1 with $k=1$.

First,

We need

Dividing by $4$ gives

Thus

Next,

Since $A=1$, the condition

becomes

so

Finally, the vertex has $x$-coordinate

We need

Combining all three conditions gives

Example 2

Find all real values of $c$ such that both roots of

$$x^2-5x+c=0$$

lie in the interval $(1,4)$.

Let

Here,

The vertex is

Since

the vertex condition is already satisfied.

Now compute the discriminant:

We need

so

Next,

and

Since $A=1$, we need

and

As such

so

Combining the conditions gives

Example 3

Find all real values of $c$ such that exactly one root of

$$x^2-5x+c=0$$

lies in the interval $(1,4)$.

Let

For exactly one root in $(1,4)$, use Case 5:

But

and

As such

This can never be negative.

Therefore, there are no real values of $c$ for which exactly one root lies in $(1,4)$.

This makes sense because the quadratic

has axis of symmetry

which is exactly in the middle of the interval. Its roots are symmetric around the centre of the interval, so they enter or leave the interval together.

Example 4

Find all real values of $c$ such that one root of

$$x^2-5x+c=0$$

lies to the left of $1$ and the other lies to the right of $4$.

We want

So we use Case 6:

and

Since $A=1$, this becomes

and

Now

and

Therefore we need

So

Therefore, for every

one root lies to the left of $1$ and the other lies to the right of $4$.

At Least One Root in an Interval

Sometimes a problem asks for at least one root in

This is not a single new case. It is the union of two possibilities:

• Exactly one root lies in the interval
• Both roots lie in the interval

So you should combine Case 5 and Case 4.

In contest problems, it is often easier to solve these two cases separately and then take the union of the answer sets.

Key Contest Insight

Root-location problems are graph problems disguised as algebra problems.

The conditions are not random inequalities. Each one has a geometric meaning:

• $\Delta\ge 0$ ensures real roots exist.
• $Af(k)>0$ means $k$ lies outside the interval between the roots.
• $Af(k)<0$ means $k$ lies between the roots.
• $-\frac{B}{2A}$ gives the horizontal position of the vertex.

Once these ideas are clear, the cases become much easier to remember.

Strategy Tips

• Start by writing down $f(x)$, $A$, $B$, $C$, and $\Delta$
• Compute the vertex coordinate $h=-\frac{B}{2A}$
• Decide which root-location case matches the problem
• Write the required inequalities carefully
• Solve all inequalities and take their common solution
• Be careful with strict intervals such as $(k_1,k_2)$
• If the problem says 'at least one root', split it into cases

Common Pitfalls

• Forgetting to check the discriminant when both roots are required
• Using $f(k)>0$ instead of $Af(k)>0$ when the parabola may open downwards
• Forgetting that the vertex lies midway between the two roots
• Combining endpoint conditions incorrectly
• Treating 'exactly one root in an interval' the same as 'both roots in an interval'
• Ignoring whether the interval is open or closed

Practice Problems