Find all real numbers $a$ for equation $x^3 + ax^2 + 51x + 2023=0$, has two equal roots.

Question

Problem:

Find all real numbers $a$ for which the equation, $x^3 + ax^2 + 51x + 2023=0$, has two equal roots.

• This problem is from an algebra round of a local high school math competition that has already ended

My Work:

To find when $x^3 + ax^2 + 51x + 2023$ has two equal roots, we can use the properties of the discriminant. For a cubic polynomial, the discriminant is given by the following formula:

$Δ = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2$

In our case, the cubic polynomial is $x^3 + ax^2 + 51x + 2023$, so $a = 1$, $b = a$, $c = 51$, and $d = 2023$.

If the polynomial has two equal roots, its discriminant must be equal to zero. So, we can set Δ equal to zero and solve for a:

$0 = 18(1)(a)(51)(2023) - 4a^3(2023) + a^2(51)^2 - 4(1)(51)^3 - 27(1)^2(2023)^2$

However, after this step, I am really stuck. I would appreciate some help! Thank you!

Answer 1

(Proceeding in a different manner as I don't see a realistic way to complete your solution in a high school competition setting.)

We have $x^3 + ax^2 + 51x + 2023 = (x+b)^2(x+c)$.
By comparing coefficients,

• $ a = 2b + c$
• $ 2bc + b^2 = 51$
• $ b^2c = 2023$

Substitute $ c = \frac{2023}{b^2}$ to eliminate it, we get $ 51 b = b(b^2 + 2bc) = b^3 + 2\times 2023$.
Facotorizing this cubic (start by testing the factors of 4046), we get $0 = (b+17) ( b^2 - 17b + 238) $.

The only real root is $b = -17$, which gives $ c = 7, a = -27$.

Answer 2

If it has two equal roots, it means all three roots are real, let the double roots at $x=c$, so

$$x^3 + ax^2 + 51x + 2023=(x-c)^2(x-b)=x^3-(2c+b)x^2+(c^2+2bc)x-bc^2$$

Compare coefficients

$$-bc^2=2023\tag{1}$$ $$c^2+2bc=51\tag{2}$$ $$b+2c=-a\tag{3}$$

From (1), (2), eliminate $b$, and we get

$$0=c^3 - 51 c - 4046=(-17 + c) (238 + 17 c + c^2)$$ Therefore, $c=17, b=-7$

From (3), we get $a=-27$

Answer 3

Simplifying that last expression makes things easier to see:

$$0 = 18(1)(a)(51)(2023) - 4a^3(2023) + a^2(51)^2 - 4(1)(51)^3 - 27(1)^2(2023)^2$$

$$=-8092 a^3 + 2601 a^2 + 1857114a -111028887 = 28a^3 - 9a^2 + 6426a + 384183$$ (the final equivalence above is achieved via a division by $-289$ to make the numbers more manageable)

Now we're just solving a straightforward cubic polynomial for $a$. Factor out an $(a+27)$ to get $$28a^3 - 9a^2 + 6426a + 384183 = (a+27)(28a^2 - 765a + 14229) = 0$$

Since the second part of that factored expression (the quadratic piece) has no real roots, we can see that there is one real value of $a$ that will solve the equation, that being $a=-27$.

Answer 4

Alt. hint: $\;$ let the double root be $u$ and the third root be $v$. Then by Vieta's relations $u^2 + 2 uv = 51$ and $u^2v = -2023$. Solve the system for $u,v$, then $a = -(2u+v)$.

Answer 5

Let $\;P(x)=x^3+ax^2+51x+2023\,.$

It results that $\;P’(x)=3x^2+2ax+51\,.$

If $\,P(x)\,$ has $2$ equal roots $\,x_1\!=\!x_2\!=\!r\,,\,$ then $\,P(r)\!=\!P’(r)\!=\!0\,,$

hence, $\;rP’(r)-2P(r)=0\;,\;$ that is

$r^3-51r-4046=(r-17)(r^2+17r+238)=0\;,$

consequently, $\;r=17\;.$

Moreover,

$P’(17)=0\;$ $\implies$ $\;918+34a=0\;$ $\implies$ $\;a=-27\,.$

Answer 6

This started out as an observation in the comments, but I should probably develop it further...

I found it of interest that $ \ 2023 \ $ can be factored as $ \ 7·289 \ = \ 7·17^2 \ $ (which would not be unreasonable to ask participants in a competition to find*), and that it seemed to me that someone took this as the "germ" for writing this problem.

$ ^{*} $ I've come to believe that it is generally a good idea for someone entering a math contest to know the prime factorization of the current year (or that it's a prime -- the next one being 2027), given the predilection of problem-posers to use that number somewhere in the exam.

So if we apply the Viete relation that the constant term in $ \ x^3 + ax^2 + 51x + 2023 \ $ is the negative of the product of the zeroes $ \ -2023 \ = \ ( \ -7 \ )·( \ \pm 17 \ )^2 \ \ , \ $ we have two possibilities for $ \ a \ $ as the negative of the sum of these zeroes, either $ \ -( \ -7 + 2·17 \ ) \ = \ -27 \ $ or $ \ -( \ -7 + 2·[-17] \ ) \ = \ 41 \ \ . \ $ If we check these against the linear coefficient, which is the sum of the pairwise-products of the zeroes, we have either $$ b \ \ = \ \ (-7)·17 \ + \ (-7)·17 \ + \ 17·17 \ \ = \ \ -238 + 289 \ \ = \ \ 51 \ \ \ \text{or} $$ $$ = \ \ (-7)·(-17) \ + \ (-7)·(-17) \ + \ (-17)·(-17) \ \ = \ \ 238 + 289 \ \ = \ \ 527 \ \ \ . $$

Only the first of these is consistent with the given polynomial, so we have $ \ a \ = \ -27 \ \ . $

In the comment, I did not intend to suggest that this was the only possible solution, but only that this integer solution was something the poser had in mind. As there were already five posted answers, I was limiting this to a remark.

However, we can go on to show that there are no other real-number solutions. As pointed out in other posts, if a cubic polynomial with real coefficients has a real "double zero", then all three zeroes must be real. The Rule of Signs tells us that

• for $ \ a \ > \ 0 \ \ , \ $ there are no positive and three or one negative real zeroes, and

• for $ \ a \ < \ 0 \ \ , \ $ there are two positive and one negative real zero.

[ EDIT (6/5) -- $ \ a \ = \ 0 \ $ is the one value that can be "eliminated" by the Rule of Signs, since $ \ x^3 + 51x + 2023 \ $ has no sign-changes and $ \ -x^3 - 51x + 2023 \ $ has but one, which is not consistent with the requirement of a real double zero.]

We see that the value for $ \ a \ $ shown above is consistent with this. Using those zeroes as the "basis" for finding any others that may exist, let us write the "double zero" as $ \ r \ = \ \frac{17}{\rho} \ $ and the third zero as $ \ s \ = \ \frac{-7}{\sigma} \ \ , \ $ with $ \ \rho \ $ and $ \ \sigma \ $ being (non-zero) real numbers. The product of the zeroes becomes $$ -c \ \ = \ \ -2023 \ \ = \ \ \ \left(\frac{17}{\rho} \right)^2·\left( \frac{-7}{\sigma} \right) \ \ \Rightarrow \ \ \rho^2·\sigma \ \ = \ \ 1 \ \ . \quad \mathbf{[ \ 1 \ ] } $$

Now the ratio $ \ \frac{b}{-c} \ $ for this monic polynomial is the ratio of the reciprocals of the zeroes, which gives us

$$ \frac{b}{-c} \ \ = \ \ \frac{51}{-2023} \ \ = \ \ -\frac{3}{119} \ \ = \ \ \frac{r^2 \ + \ 2rs}{r^2·s} \ \ = \ \ \frac{2}{r} \ + \ \frac{1}{s} $$ $$ = \ \ \frac{2·\rho}{17} \ + \ \frac{\sigma}{-7} \ \ = \ \ \frac{-14·\rho \ + \ 17·\sigma}{-119} \ \ \Rightarrow \ \ 14·\rho \ - \ 17·\sigma \ \ = \ \ -3 \ \ . \quad \mathbf{[ \ 2 \ ] } $$

[ EDIT (6/5) -- Here also, we can "reject" $ \ a \ = \ 0 \ \ , \ $ as this would imply $ \ -(2r + s) \ = \ 0 \ \Rightarrow \ s \ = \ -2r \ \Rightarrow \ b \ = \ r^2 + 2rs \ = \ -3r^2 \ = \ 51 \ \ , $ $ c \ = \ r^2·s \ = \ -2r^3 \ = \ -2023 \ \ , \ $ the first of these results not permissible, since $ \ r \ $ must be real, and the two conclusions not being mutually consistent.]

Solving these two equations simultaneously produces $$ \sigma \ \ = \ \ \frac{1}{\rho^2} \ \ \rightarrow \ \ 14·\rho \ - \ 17·\frac{1}{\rho^2} \ \ = \ \ -3 \ \ \Rightarrow \ \ 14·\rho^3 \ + \ 3·\rho^2 \ - \ 17 \ \ = \ \ 0 \ \ , $$ for which we know (or can readily "eyeball" that) $ \ \rho \ = \ 1 \ $ is a solution. From this we have $ \ 14·\rho^3 + 3·\rho^2 - 17 \ = \ (\rho - 1)·(14·\rho^2 + 17·\rho + 17) \ \ , \ $ with the quadratic factor having a negative discriminant $ \ \Delta \ = \ 17^2 - 4·14·17 \ \ . $

So $ \ \rho \ = \ 1 \ \ , \ \ \sigma \ = \ 1 \ $ is the only real-valued solution for the pair of equations, making $ \ (x + 7)·(x - 17)^2 \ \ $ the sole factorization of our cubic polynomial. Hence, $ \ a \ = \ -27 \ $ is indeed the only real-number value for this problem.

[This system also yields two complex-number solutions:

$$ \rho \ \ = \ \ -\frac{17}{28} \ \pm \ i·\frac{\sqrt{663}}{28} \ \ \Rightarrow \ \ r \ \ = \ \ -\frac{17}{2} \ \mp \ i·\frac{\sqrt{663}}{2} \ \ , $$ $$ \sigma \ \ = \ \ -\frac{11}{34} \ \pm \ i·\sqrt{\frac{39}{17}} \ \ \Rightarrow \ \ s \ \ = \ \ \frac{187}{56} \ \pm \ i·\frac{17·\sqrt{663}}{56} $$

$$ \Rightarrow \ \ -r^2·s \ \ = \ \ c \ \ = \ \ 2023 \ \ \ , \ \ \ r^2 \ + \ 2·r·s \ \ = \ \ b \ \ = \ \ 51 \ \ \ , $$ $$ a \ \ = \ \ -( \ 2r \ + \ s \ ) \ \ = \ \ -\frac{785}{56} \ \mp \ i·\frac{39·\sqrt{663}}{56} \ \ . \ ] $$