When you solve a given equation for a particular variable, you are finding what is known as the root of the equation. A root of the function f (x) is a value of x for which f (x) = 0. The graph corresponding to y = f (x) will cross the X-axis at points corresponding to the location of roots of the function.
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Consider the equation y = (x + 3) (x-2).
What are the roots of the above equation?
Solution
The roots of the above equation are values of x for which y = 0.
Therefore,
y = (x + 3) (x-2) = 0
when
x + 3 = 0, \(\rightarrow\) x = -3
or, x - 2 = 0 \(\rightarrow\) x = 2
Thus, the roots of the given equation are x = 2 and x = -3.
Consider the following equation:
y = (x-2) (x + 4) (x-6)
The following graph shows the corresponding curve.
Graph for the function y = (x-2) (x + 4) (x-6)
From the above graph, we can see that the points at which the curve intersects the X-axis are at x = -4, x = 2 and x = 6. Thus, the roots of the given equation are -4, 2 and 6.
Now look at the points on the graph marked A (corresponding to x = 1) and B (corresponding to x = 4). A lies above the X-axis, and B lies below the X-axis. Given that the graph is continuous between A and B (there is an unbroken line connecting A and B on the graph), this implies that there necessarily has to be at least one root between A and B. For the curve to travel from above the X-axis to below it, it has to cross the X-axis at some point.
The graph being continuous in the interval between A and B is a necessary condition here. If the graph were discontinuous, there would be no compulsion for the line to intersect the X-axis. For example, you could have a function that diverges at a vertical asymptote in the given interval.
We can generalise the above discussion to obtain the location of roots theorem:
If the function f (x) is continuous in the interval [a, b] and f (a) and f (b) have opposite signs, then f (x) has at least one root, x, that lies between a and b, i.e. a < x < b.
Satisfaction of the above condition means that there is at least one root between \(\x = a\) and \(\x = b\). However, it does not necessarily mean that there is only one root between \(\x = a\) and \(\x = b\). For example, consider points C and D in the above graph. C and D satisfy the condition that they are of opposite signs (the value of the function is positive at C and negative at D), but we can see from the graph that there are three roots between C and D and not just one.
Conversely, just because two points lie on the same side of the X-axis (i.e. the value of f (x) has the same sign for two values of x) does not necessarily mean that there are no roots between them. Consider points A and C on the graph. Both are above the X-axis, i.e. the value of f (x) is positive at both points. However, we see that there are two distinct roots (at \(\x = 2\) and \(\x = 6\)) between these two points.
The application of the location of roots theorem cannot be used directly to find the exact root(s) of a function. However, it can be very useful for estimating the approximate location of the roots of a function. In many methods, the location of roots theorem is applied to find an initial approximation of the roots of a function. The successive application of the theorem is used to iteratively get closer to the function's root(s). Check out our article on Iterative Methods for more details.
In the following section, we will solve some example problems on the location of roots.
Example 1
Show that the function f(x) = x³ - x + 5 has at least one root between x = -2 and x = -1
Solution 1
f(-2) = -2³ - (-2) + 5 = -1
f(-1) = -1³ - (-1) + 5 = 5
Since f(-2) is negative, and f(-1) is positive, according to the Location of Roots theorem, this implies that there is at least one root of f(x) between -2 and -1.
Example 2
Given f (x) = x³ - 4x² + 3x + 1, show that f (x) has a root between 1.4 and 1.5
Solution 2
f (1.4) = 1.4³ - 4 x 1.4² + 3 x 1.4 + 1 = 0.104
f (1.5) = 1.5³ - 4 x 1.5² + 3 x 1.5 + 1 = -0.125
Since f(1.4) is positive, and f(1.5) is negative, according to the Location of Roots theorem, this implies that there is at least one root of f(x) between 1.4 and 1.5.
Example 3
For a quadratic function f(x), f(2) = 3.6, f(3) = -2.2, f(4) = -0.1, f(5) = 0.9.
From the above information can we conclude if there is a root between
a) 2 and 3 ?
b) 3 and 4 ?
c) 4 and 5 ?
Solution 3
a) We see that there is a change of sign between f(2) (positive) and f(3) (negative). So we can say that there is at least one root of f(x) between 2 and 3.
b) We see that there is no change of sign between f(3) (negative) and f(4) (negative). If there has to be a root between f(3) and f(4), there would need to be at least two roots since the sign would have to change to positive and then back to negative. But we know that quadratic equations have at most two roots, and we have already found a different location for one root. This means that there is no root of f(x) between 3 and 4.
c) We see that there is a change of sign between f(4) (negative) and f(5) (positive). So we can say that the remaining root of the quadratic, f(x) lies between 4 and 5.
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How do you find the position of roots?
If the function f(x) is continuous between a and b, and f(a) and f(b) have opposite signs, then f(x) has at least one root, x, that lies between a and b.
What is the location of roots theorem?
If the function f(x) is continuous in the interval [a,b] and f(a) and f(b) have opposite signs, then f(x) has at least one root, x, that lies between a and b ie. a < x < b.
What is distinct root?
If all roots of the equation, are different or unequal, they are called distinct roots.
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