Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am given the following function $f : \mathbb R \mapsto \mathbb R, f(x) = x^4 - 4x + p\ \ \ $ and am asked to find $p$ such that $f$ has two identical real roots.

The proposed solution is to get the root from the relation $f(a) = f'(a) = 0$. What I'm curious about is when this relation is valid. It obviously isn't always valid (linear functions for example).

share|improve this question
    
The relation is $f(a)=f'(a)=0$. –  André Nicolas Apr 19 '11 at 18:58
    
Ah yes, thank you for the correction. –  Paul Manta Apr 19 '11 at 18:59

3 Answers 3

up vote 3 down vote accepted

If $f$ has a double root at $x=a$, then $f(x)=(x-a)^2 g(x)$.

share|improve this answer
    
Obviosuly, but I don't see how that relates to $f'(x)$. :| –  Paul Manta Apr 19 '11 at 18:55
4  
Then $f'(x)=2(x-a)g(x)+(x-a)^2g'(x)$ which has a root at $x=a$ –  Ross Millikan Apr 19 '11 at 19:03
    
Thanks! So that relation is valid when the degree of the function is greater or equal to 2. –  Paul Manta Apr 19 '11 at 19:14
2  
@Paul, you mean when the multiplicity of the root is greater than or equal to 2. –  Ross Millikan Apr 19 '11 at 21:56

The question is already answered, but I wanted to add in my intuition for the why of the solution: The degree of a polynomial determines how many roots it can have. If we were to plot the function, the real-valued roots are places where the curve crosses zero on the y-axis. If the plot has a 'hump' that doesn't cross the $y=0$ axis, then the function has at least some imaginary roots. However, if the function just touches the $y=0$ axis, then both roots for that bump are in the same place: zero. When a bump on a curve only just brushes up against the axis, then it's tangent to that line at that point. Therefore, its first derivative is also equal to the slope of that line. Consequently, there would be two solutions to your problem if you were to solve it for this graph:

Fourth order polynomial plot

You could translate the graph downward so that the middle bump was tangent with $y=0$, or you could translate the plot upward so that the left and right bumps were.

share|improve this answer

The question is already answered (twice). It is however perhaps useful to make a comment about polynomials of degree less than $2$, since the poster seems to believe they are an exception. Such polynomials are certainly uninteresting in this context, but they are not an exception.

Below, we will tacitly assume that we are dealing with polynomial with real coefficients, though with the appropriate algebraic definition of the derivative, one can prove a more general result.

Theorem: Let $P(x)$ be a polynomial. Then the real number $a$ is an (at least) double root of $P(x)$ if and only if $P(a)=P'(a)=0$.

(The "at least" is there because for example $(x-1)^3$ has a triple root at $x=1$. There are two possible interpretations of the term "double root": multiplicity at least $2$ or multiplicity exactly $2$. I am just being careful.)

The theorem holds for all polynomials, without exception. For example, let $P(x)=x-17$. It is true that for any $a$, if $a$ is a double root of $P(x)$, then $P(a)=P'(a)=0$, for $P(x)$ has no double roots. And it is true that if $P(a)=P'(a)=0$, then $a$ is a double root of $P(x)$, since in fact there is no $a$ such that $P(a)=P'(a)=0$.

The same remark would hold for polynomials of degree $0$ (non-zero constants). Finally, let us look at the $0$ polynomial, which in algebra is usually said to have no degree, or degree $-1$, or degree $-\infty$. For this trivial polynomial, every $a$ is a root of multiplicity at least $2$, indeed of infinite multiplicity. And it is true that for every $a$, $P(a)=P'(a)=0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.