# Finding double root. An easier way?

Given the polynomial $f = X^4 - 6X^3 + 13X^2 + aX + b$ you have to find the values of $a$ and $b$ such that $f$ has two double roots.

I went about this by writing the polynomial as: $$f = X^4 - 6X^3 + 13X^2 + aX + b = (X - r)^2(X - s)^2$$

... and from there I arrived at a system of four equations from where I could find the desired values. However, there's quite a bit of work to do it this way; is there a shortcut?

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a=b and x=0 seems to work? Or do you want a full specification? It might be helpful to note that then $f=0$ and $f'=0$ if a point is a double root –  Thomas Rot May 10 '11 at 15:20
Hint: Any multiple root is also a root of the derivative. Divide $f$ by $f'$ using ordinary polynomial "long division." The remainder should tell you everything. –  André Nicolas May 10 '11 at 15:21
@user6312, I could not carry out that hint, does it really work in this case? –  quanta May 11 '11 at 13:54

## 4 Answers

HINT $\$ Simpler, first solve $\rm\ f\: =\: (x^2 + c\ x + d)^2\:.\:$ This yields

$$\rm x^4 - 6\ x^3 +13\ x^2 +\cdots\ =\ x^4 + 2\:c\ x^3 + (2\:d+c^2)\ x^2\:+\cdots$$

hence $\rm\ 2\:c = -6\$ so $\rm\ c = -3\:.\$ And $\rm\ 2\:d+c^2 = 2\:d+9 = 13\$ so $\rm\ d = 2\:.$

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Don't forget to add or check the restriction that the quadratic has two real zeros. –  lhf May 10 '11 at 15:40
@lhf That plays not role when you proceed as I hinted. –  Bill Dubuque May 10 '11 at 15:47
yes, for that specific polynomial. –  lhf May 10 '11 at 15:52

Both sides are identical polynomial expressions. The roots of the right are r,r,s,s. The sum of the roots is 6; this is seen by looking at the left. Hence, $2r+2s = 6$ so $r + s = 3$. The coefficient of the squared term is is the sum of the roots multiplied two at a time. From the left we see it is 13. From the point of view of $r$ and $s$, it is

$$r^2 + s^2 + 4rs.$$

Now substitute in $s = 3 - r$ and solve. You get $r = 1$ or $r = -2$. Now check out both possibilities.

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You could apply Vieta's theorem; you would also arrive at a 4x4 non-linear system of equations, thus I'm not sure whether this method is lots easier than yours. The system of equations would look like this

$$\begin{vmatrix}2 x_1 + 2 x_2 = 6\\ x_1^2 + 4 x_1 x_2 + x_2^2 = 13\\ 2 x_1^2 x_2 + 2 x_1 x_2^2 = -a\\ x_1^2 x_2^2 = b \end{vmatrix}$$

and will give the solution $a = -12$ and $b=4$.

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Interesting, I didn't know this principle had a name. –  ncmathsadist May 10 '11 at 15:57

To elaborate on the calculus based method (which, in comparison to the other solutions here, I do not recommend), if we let $f=X^4−6X^3+13X^2+aX+b$, then we have double roots at $r$ and $s$ if $GCD(f,f')=k(x-r)(x-s)$ for some non-zero constant $k$. Going through the Euclidean algorithm to find $GCD(f,f')$, the first remainder we get is (up to a factor) $2X^2-6X(13+a)-(3a+8b)$, which cannot be $0$ for any choice of $a$ and $b$. Since we need that the greatest common divisor be quadratic, the next remainder has to be zero. Proceeding, the next remainder is $(104a+9a^2 +276b + 24ab)+2x(1352+222a+9a^2+4b)$. Therefore, we must have that both coefficients are zero.

Since $b$ appears linearly in both coefficients, we can solve for $b$ and set both expressions equal to each other to yield that $-b=\frac{1}{4}(9a^2+222a+1352)=\frac{9a^2+104a}{12(23+2a)}$. This gives a cubic equation for $a$, and once solved, yields $b$ too.

This can in principle be solved using Cardano's formula, or otherwise, and has the "advantage" that it does not require the intermediate step of writing things in terms of the roots. On the other hand, dealing with cubics is a huge pain.

A hybrid of this method and the other method might be to use the fact that, if the roots of $f$ are $r$ and $s$ (each with multiplicity $2$), then $2X^2−6X(13+a)−(3a+8b)$ has roots $r$ and $s$. We then have that $r+r+s+s=6$ and $r+s=3(13+a)$, so $a=-12$.

Then, the equation $1352+222a+9a^2+4b=0$, simplifies to $-16+4b=0$, or $b=4$.

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What is $(104a+9a^2 +276b + 24ab)+2x(1352+222a+9a^2+4b)$ in terms of $2X^2-6X(13+a)-(3a+8b)$, $f$ and $f'$? By the way thanks for posting this because I was wondering how to use this technique. –  quanta May 11 '11 at 17:50
@quanta: When you're doing the Euclidean algorithm, starting with $f$ and $f'$, $2X^2−6X(13+a)−(3a+8b)$ is the (8 times) remainder of dividing $f$ by $f'$, and $(104a+9a2+276b+24ab)+2x(1352+222a+9a2+4b)$ is the remainder of dividing $f'$ by $2X^2−6X(13+a)−(3a+8b)$, though I might have divided the remainder by $2$ to get slightly smaller numbers. If you're asking for the remainderless quotients, I don't have that information on me, but I'm sure that wolframalpha can help you figure it out. Use the PolynomialQuotient[p,q,x] and PolynomialRemainder[p,q,x] functions. –  Aaron May 11 '11 at 18:36