Root placement in an interval: discriminants, vertices, and endpoints

Turn a root question into an intersection question

A root-placement problem asks how many real roots of an equation lie in a specified interval. For a quadratic, three pieces of information are especially useful:

• The discriminant: are there two distinct roots, a double root, or no real roots?
• The vertex: where is the axis relative to the interval?
• The endpoint values: on which side of the horizontal axis is the graph at the boundaries?

We will assume the quadratic opens upward. If its leading coefficient is negative, multiply the equation by minus one first. This leaves its roots unchanged and makes the sign criteria below applicable.

The quadratic formula is valid too, but directly solving several radical inequalities is often longer than using the graph's structure.

Two distinct roots in an open interval

For $f(x)=Ax^2+Bx+C$ with $A>0$, let $v=-B/(2A)$ and $D=B^2-4AC$. Both roots lie in $(\alpha,\beta)$ exactly when

$$\boxed{D>0,\quad\alpha<v<\beta,\quad f(\alpha)>0,\quad f(\beta)>0.}$$

The vertex is below the axis because $D>0$. Positive endpoint values and a vertex inside force one crossing on each side of it before reaching the endpoints.

For a closed interval, allowing the roots at the endpoints changes the endpoint inequalities to nonstrict ones. Keeping $D>0$ still distinguishes two roots from a double root.

Problem 1: two roots in (-1, 2)

Problem. Find all real $a$ for which

$$x^2+ax+a^2-3=0$$

has two distinct roots in $-1<x<2$.

For $f(x)=x^2+ax+a^2-3$,

$$v=-\frac a2,\quad D=12-3a^2,\quad f(-1)=(a-2)(a+1),\quad f(2)=(a+1)^2.$$

The four conditions become

$$-2<a<2,\quad-4<a<2,\quad a<-1\text{ or }a>2,\quad a\ne-1.$$

Their intersection is

$$\boxed{-2<a<-1.}$$

At $a=-2$ the discriminant is zero, so the two roots merge. At $a=-1$ one root is an excluded endpoint. Both strict boundaries are therefore necessary.

Exactly one root in an open interval

There are three possibilities to examine:

1. The endpoint values have opposite signs, giving one crossing inside.
2. A double root lies strictly inside the interval.
3. An endpoint is itself a root, and the other root lies strictly inside.

The third case is easily missed because an endpoint of an open interval is excluded, while the other root may still count.

Problem 2: one distinct root, including a double root

Problem. Find all real $a$ for which

$$x^2+2ax-a+1=0$$

has exactly one distinct root in $-1<x<3$. A double root counts as one.

Here $v=-a$, $D=4(a^2+a-1)$, $f(-1)=2-3a$, and $f(3)=10+5a$.

First handle endpoint roots:

• At $a=2/3$, the roots are $-1$ and $-1/3$, so exactly one is inside.
• At $a=-2$, the roots are $1$ and $3$, again leaving exactly one inside.

Away from those parameter values, opposite endpoint signs require

$$(2-3a)(10+5a)<0,$$

which gives $a<-2$ or $a>2/3$.

The double-root case requires $a^2+a-1=0$. Its two values $(-1\pm\sqrt5)/2$ both place $v=-a$ inside $(-1,3)$.

Combining the cases,

$$\boxed{a\le-2\quad\text{or}\quad a=\frac{-1\pm\sqrt5}{2}\quad\text{or}\quad a\ge\frac23.}$$

The same problem can be approached by separating the parameter and rotating a line.

At least one root in a closed interval

For an upward quadratic, a useful exhaustive criterion on $[\alpha,\beta]$ is

$$f(\alpha)f(\beta)\le0$$

or, simultaneously,

$$D\ge0,\quad\alpha\le v\le\beta,\quad f(\alpha)\ge0,\quad f(\beta)\ge0.$$

The first case includes an endpoint root and opposite-sign endpoints. The second includes a vertex at or below the axis while both endpoints are at or above it. The cases can overlap, but together they omit no valid configuration.

Equivalently, the continuous function must have a minimum at most zero and a maximum at least zero on the interval. The maximum of an upward quadratic is at an endpoint; the minimum is at the vertex if it lies inside, or at an endpoint otherwise.

Problem 3: a closed interval

Problem. Find all real $a$ for which

$$x^2+ax+a^2-3=0$$

has at least one root in $-1\le x\le3/2$.

The discriminant first requires $-2\le a\le2$. For every $a$ in this range, the vertex $-a/2$ lies in $[-1,1]$, and therefore in the required interval. Its minimum value is nonpositive. Thus a root exists in the interval precisely when at least one endpoint value is nonnegative:

$$f(-1)=a^2-a-2\ge0\quad\text{or}\quad f(3/2)=a^2+\frac32a-\frac34\ge0.$$

The first inequality gives $a\le-1$ or $a\ge2$. The second gives

$$a\le\frac{-3-\sqrt{21}}4\quad\text{or}\quad a\ge\frac{-3+\sqrt{21}}4.$$

Intersecting their union with $[-2,2]$ yields

$$\boxed{-2\le a\le-1\quad\text{or}\quad\frac{-3+\sqrt{21}}4\le a\le2.}$$

The endpoints are included because roots at the interval boundaries are allowed.

Problem 4: a cubic and a separated constant

Problem. Find all real $a$ for which

$$x^3-3x-a=0$$

has a root in $0<x<1/2$.

Write $x^3-3x=a$. The fixed function $h(x)=x^3-3x$ has derivative $3x^2-3<0$ throughout $(0,1/2)$, so it is strictly decreasing there. Its endpoint limits are

$$h(0)=0,\qquad h(1/2)=\frac18-\frac32=-\frac{11}8.$$

Continuity and strict decrease show that its image is exactly $(-11/8,0)$. Therefore

$$\boxed{-\frac{11}8<a<0.}$$

For higher-degree equations, parameter separation can simplify the question to the range or intersection count of a fixed function. It is a useful choice rather than a rule forbidding other approaches; derivatives, factorization, or explicit roots can also be appropriate.

The endpoint habit

Write down whether each interval endpoint is included before starting. Then test critical parameter values directly in the original equation. A discriminant equality, a tangency, and a root crossing an endpoint are different events, and each can change a root count.