The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.

Question

The given question is:

The equation $x^{4}-3x^{3}-6x^{2}+ax+b=0$ has a triple root. Find $a$ and $b$, and hence all roots of this equation.

I am confused about how to work out this question, but I feel like it has something to do with using α,β,γ, δ as roots of the quartic.

I would greatly appreciate all help in a simple and efficient method to solving this question. Please note, according to the answers, there is only one value of $a$ and one value of $b$.

Answer 1

If a polynomial has a $n$-th order repeated root, then that root will also be be a root of the first to the $(n-1)$th derivatives of the polynomial.

So the triple root will also be a root of the first and second derivative, the latter is $12x^2 - 18x - 12$. This quadratic has roots of $2, -\frac 12$.

Based on the first derivative, which is $4x^3 - 9x^2 - 12x + a$, you get $a = 28$ or $a = -\frac {13}4$. The corresponding $b$ values from the original polynomial are $b = -24$ and $b = -\frac 9{16}$ respectively. So $(a,b) = (28,-24)$ or $(-\frac{13}{4}, -\frac {9}{16})$.

If the question stated that $(a,b)$ were integers, then you could discard one set of solutions. But as it stands, both are admissible.

Answer 2

We know that the equation has a triple root, so if let $\alpha$ be the triple root and $\beta$ be the single root, we get that \begin{equation} x^4 - 3x^3-6x^2 + ax + b = (x-\alpha)^3(x-\beta) \end{equation} By expanding the RHS, we see that \begin{equation} x^4 - 3x^3-6x^2 + ax + b = x^4 - (3\alpha+\beta)x^3 + (3\alpha\beta + 3\alpha^2)x^2 + \cdots \end{equation} Now equating the coefficients of the $x^2$ and $x^3$ term, we have the simultaneous equations \begin{equation} 3 = 3\alpha+\beta \quad ; \quad -6 = 3\alpha\beta + 3\alpha^2 \end{equation} By a simple substitution we conclude $(\alpha,\beta) = (2,-3)$ or $(\alpha,\beta) = (-1/2, 9/2)$. We can check that these indeed satisfy our conditions. It then follows that we have either $(a,b) = (28,-24)$ or $(a,b) = (-13/4,-9/16)$.

Answer 3

the symmetric polynomials of the roots form the coefficients: multiplying out $\quad (x-r_1)(x-r_2)(x-r_3)(x-r_4)=0\quad$ you see that the second coefficient is the negative sum of the roots.

And we're given that$\quad r_1=r_2=r_3\quad$ and some other independent $\quad r_4$:

so $\quad 3r_1+r_4=3 \iff r_4=3-3r_1\quad$ and the third coefficient is the six term product-sum: $$r_1r_2+r_1r_3+r_1r_4+r_2r_3+r_2r_4+r_3r_4=r_1^2+r_1^2+r_1r_4+r_1^2+r_1r_4+r_1r_4\\=3r_1^2+3r_1r_4=-6\iff 3r_1(r_1+r_4)=-6 \iff r_1(r_1+3-3r_1)=-2\\\iff -2r_1^2+3r_1+2=0 \iff (2r_1+1)(r_1-2)=0$$ it's everything you need, the rest is computation. Then $\quad -a,\quad$ the negative fourth coefficient is $$r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4$$ and of course the final coefficient is $\quad r_1r_2r_3r_4$

Answer 4

Working from the Viete relations is probably the most direct way to obtain the zeroes of the polynomials, and thence the unknown coefficients. We might also approach the problem without the use of calculus in this way.

If we are to take it that all of the coefficients of $ \ f(x) \ = \ x^4 - 3x^3 - 6x^2 + ax + b \ $ are real, then the statement that the polynomial possesses a triple zero $ \ r \ $ requires that all of the zeroes are real, since any complex zeroes would have to occur in "conjugate pairs". (We will call the fourth real zero $ \ s \ \ . \ ) $ We next infer the following from the Rule of Signs. There can be no more than three "sign-changes" among the coefficients, since there are two consecutive negative coefficients shown for $ \ f(x) \ $ and the first two coefficients of $ \ f(-x) \ $ are positive. If $ \ r \ > \ 0 \ \ , \ $ three sign-changes are only possible with $ \ a \ $ positive and $ \ b \ $ negative; for $ \ f(-x) \ = \ x^4 + 3x^3 - 6x^2 - ax + b \ \ , $ there can only be three sign-changes with both $ \ a \ $ and $ \ b \ $ being negative. So it must be the case that $ \ f(0) \ = \ b \ < \ 0 \ \ . $

The Viete relations tell us that the constant term $ \ b \ $ in a monic polynomial of even degree is the product of its roots, so $ \ b \ = \ r^3·s \ \ . \ $ That $ \ b \ $ is negative then requires that $ \ r \ $ and $ \ s \ $ are of opposite signs.

This raises the interesting possibility that there are two solutions for $ \ f(x) \ = \ (x - r)^3·(x - s) \ \ , \ $ the curve for one of these being a "reflection" of the other about a vertical line $ \ x \ = \ c \ \ . \ $ In terms of function transformations, this corresponds to a reflection about the $ \ y-$axis and a "horizontal shift" by $ \ 2c \ $ units, producing $ \ f(2c - x) \ = \ (2c - x)^4 - 3·(2c - x)^3 - 6·(2c - x)^2 + a·(2c - x) + b \ \ . \ $ Since this is to represent a second solution for $ \ f(x) \ \ , \ $ we can "match up" the cubic and quadratic terms to find that $ \ (3 - 8c)·x^3 \ = \ -3x^3 \ $ and $ \ (24c^2 - 18c - 6)·x^2 \ = \ -6x^2 \ \ , \ $ the two equations together implying $ \ c \ = \ \frac34 \ \ . \ $ [The graph presented at the end summarizes the results found in this discussion.]

We now have two functions, $ \ f_1(x) \ = \ (x - r)^3·(x - s) \ $ and $ \ f_2(x) \ = \ (x - r')^3·(x - s') \ \ , \ $ with $ \ r', s \ < \ 0 \ < \ r , s' \ \ , \ $ which have the same cubic and quadratic coefficients; this permits us to write $ \ 3r + s \ = \ 3r' + s' \ $ and $ \ 3r^2 + 3rs \ = \ 3r'^2 + 3r's' \ \ . \ $ As we have established that $ \ r' \ = \ 2c - r \ = \ \frac32 - r \ $ and, similarly, $ \ s' \ = \ \frac32 - s \ \ , \ $ we obtain $$ 3r \ + \ s \ \ = \ \ 3r' \ + \ s' \ \ \Rightarrow \ \ 3r \ + \ s \ \ = \ \ 3·\left(\frac32 - r \right) \ + \ \left(\frac32 - s \right) \ \ \Rightarrow \ \ 6r \ + \ 2s \ \ = \ \ 6 $$ $$ \Rightarrow \ \ s \ \ = \ \ 3 \ - \ 3r \ \ ; $$ Simplifying $ \ 3r^2 \ + \ 3rs \ = \ 3·(\frac32 - r)^2 \ + \ 3·(\frac32 - r)·(\frac32 - s) \ $ produces the same result for $ \ s \ $ , so we obtain no additional information from this. (This produces less directly what Jason Choy and MaximusFastidiousIrreverence find from the Viete relations.)

We now have enough information to determine the unknown coefficients. Using our factorization, we obtain $$ ( x - r )^3·(x - [3 - 3r]) \ \ = \ \ x^4 \ - \ 3x^3 \ - \ (6r^2 - 9r)·x^2 \ + \ (8r^3 - 9r^2)·x \ + \ (3r^3 - 3r^4) $$ $$ = \ \ x^4 \ - \ 3x^3 \ - \ 6x^2 \ + \ ax \ + \ b $$ $$ \Rightarrow \ \ 6r^2 - 9r \ \ = \ \ 6 \ \ \Rightarrow \ \ 3·(2r + 1)·(r - 2) \ \ = \ \ 0 $$ (by matching quadratic coefficients).

This leads us to the two possible values for the triple zero, and so to the coefficients of the related quartic polynomials:

$$ r \ = \ 2 \ \ \Rightarrow \ \ s \ = \ 3·(1 - 2) \ = \ -3 $$ $$ \Rightarrow \ \ a \ = \ r^2·(8r - 9) \ = \ 2^2·(8·2 - 9) \ = \ 28 \ \ , \ \ b \ = \ 3r^3·(1 - r) \ = \ 3·2^3·(1 - 2) \ = \ -24 $$ $$ \Rightarrow \ \ f_1(x) \ \ = \ \ (x - 2)^3·(x + 3) \ \ = \ \ x^4 \ - \ 3x^3 \ - \ 6x^2 \ + \ 28x \ - \ 24 \ \ \ ; $$

$$ r' \ = \ -\frac12 \ \ \Rightarrow \ \ s' \ = \ 3· \left(1 - \left[-\frac12 \right] \right) \ = \ +\frac92 $$ (in agreement with our earlier result $ \ r' \ = \ \frac32 - r \ \ , \ \ s' \ = \ \frac32 - s \ \ ) $

$$ \Rightarrow \ \ a' \ = \ \left(-\frac12 \right)^2· \left(8·\left[-\frac12 \right] - 9 \right) \ = \ -\frac{13}{4} \ \ , \ \ b' \ = \ 3·\left(-\frac12 \right)^3· \left(1 - \left[-\frac12 \right] \right) \ = \ -\frac{9}{16} $$ $$ \Rightarrow \ \ f_2(x) \ \ = \ \ \left(x + \frac12 \right)^3·\left(x - \frac92 \right) \ \ = \ \ x^4 \ - \ 3x^3 \ - \ 6x^2 \ - \ \frac{13}{4}x \ - \ \frac{9}{16} $$ (conformaing our early determination that $ \ b \ < \ 0 \ $ for either polynomial).

[The claim in the given answer that there is only one solution apparently only considers the integer values for the roots and coefficients.]

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A somewhat more laborious alternative is to utilize the distinct information we have available from the $ \ y-$intercepts of our two functions. We have $ \ f_1(x) \ = \ x^4 - 3x^3 - 6x^2 + ax + b \ $ and $ \ f_2(x) \ = \ f_1 \left(\frac32 - x \right) \ = \ x^4 - 3x^3 - 6x^2 + a'x + b' \ \ , \ $ from which we obtain $$ f_1(0) \ \ = \ \ b \ \ = \ \ 3r^3 \ - \ 3r^4 \ \ = \ \ f_2 \left(\frac32 \right) \ \ = \ \ \frac32·a' \ + \ b' \ - \ \frac{297}{16} \ \ . $$

For the linear coefficient and constant term, we have $$ a' \ \ = \ \ -( \ r'^3 \ - \ 3r'^2s' \ ) \ \ = \ \ 8r'^3 \ - \ 9r'^2 \ \ = \ \ -8r^3 \ + \ 27r^2 \ - \ 27r \ + \ \frac{27}{4} \ \ , $$ $$ b' \ \ = \ \ r'^3s' \ \ = \ \ 3·( \ r'^3 \ - \ r'^4 \ ) \ \ = \ \ 3· \left( \ -r^4 \ + \ 5r^3 \ - \ 9r^2 \ + \ \frac{27}{4}r \ - \ \frac{27}{16} \ \right) \ \ . $$

Thus, $$ 3r^3 \ - \ 3r^4 \ \ = \ \ \left( \ -12r^3 \ + \ \frac{81}{2}r^2 \ - \ \frac{81}{2}r \ + \ \frac{81}{8} \ \right) $$ $$ + \ \left( \ -3r^4 \ + \ 15r^3 \ - \ 27r^2 \ + \ \frac{81}{4}r \ - \ \frac{81}{16} \ \right) \ - \ \frac{297}{16} $$ $$ \Rightarrow \ \ \frac{27}{2}·r^2 \ - \ \frac{81}{4}·r \ - \ \frac{27}{2} \ \ = \ \ \frac{27}{4}·(2r + 1)·(r - 2) \ \ = \ \ 0 \ \ , $$ producing the same results for $ \ r \ $ and $ \ r' \ $ as above, along with all of the associated conclusions.

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