## Finding the Roots of Cubic

1. The problem statement, all variables and given/known data

Find an integer c such that the equation 4x^3 + cx - 27 = 0 has a double root.

2. Relevant equations

Ax^3+Bx^2+Cx+K = 0
Sum of Roots = -B/A
Product of Roots = (-1)^n * k/a
etc.

3. The attempt at a solution

I tried using P/Q with synthetic division to find a quadratic for the problem but I couldn't find a way to get rid of or solve for C.

I also tried manipulating the Sum of Roots and Product of Roots (both listed above) relationships to try and solve for C. No success here.

I have a hunch that it has to deal with the Sum of Roots & Product of Roots relationships, since I know what A is (leading coefficient) and I know what K is (-27, given).

I just can't seem to put it together.

The answer of the problem is C = -27, however I would like to know how to get this.

Thank you very much.
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Recognitions:
Homework Help
 Quote by TbbZz 1. The problem statement, all variables and given/known data Find an integer c such that the equation 4x^3 + cx - 27 = 0 has a double root. 2. Relevant equations Ax^3+Bx^2+Cx+K = 0 Sum of Roots = -B/A Product of Roots = (-1)^n * k/a etc.
Since this is a cubic polynomial, we know it has three (complex) zeroes. We are told that one is a double root (that is, one of the zeroes has a multiplicity of two), so there are really only two *distinct* zeroes. [Bonus question: why does this now tell us that the zeroes must all be real?] To spare ourselves having to use subscripts, let's call the "double root" R and the other root r.

You are given a couple of very nice formulas for the sum and product of the roots. What are their values? How would you express the sum and product of the roots (or zeroes) in terms of R and r? You now have two equations in two unknowns. That's your cue... Knowing the roots should let you find the missing coefficient.
 Recognitions: Gold Member Science Advisor Staff Emeritus Any cubic polynomial with a double (but not triple) root and leading coefficient 4, can be written 4(x-a)2(x-b)= 4x3-(8a+4b)x2+(4a2+8ab)x- a2b. Since you are told that the polynomial is 4x3+ cx- 27, you know that 8a+4b= 0, 4a2+ 8ab= c, and a2b= 27. That gives you three equations to solve for a, b, c. dynamicsolo, I had to think for a minute to see why all the roots must be real!

## Finding the Roots of Cubic

 Quote by dynamicsolo Since this is a cubic polynomial, we know it has three (complex) zeroes. We are told that one is a double root (that is, one of the zeroes has a multiplicity of two), so there are really only two *distinct* zeroes. [Bonus question: why does this now tell us that the zeroes must all be real?] To spare ourselves having to use subscripts, let's call the "double root" R and the other root r. You are given a couple of very nice formulas for the sum and product of the roots. What are their values? How would you express the sum and product of the roots (or zeroes) in terms of R and r? You now have two equations in two unknowns. That's your cue... Knowing the roots should let you find the missing coefficient.
I do understand that there must be three distinct zeroes. This tells me that the zeroes must all be real since every imaginary root (a+bi) means there is another imaginary root (a-bi). Because there is a double root, it would not be possible to have an imaginary root.

I worked for about 15 minutes on substituting R for the double root and r for the third root. However, I still can't seem to figure out how to solve for the missing coefficient.

I know that:

2R + r = -C/4 <-- sum of roots
2Rr = 27/4 <-- product of roots

But I'm not sure what to do next.

P.S. HallsofIvy, I'm sure your method works, but unfortunately I don't understand the concept of it, so I'd like to learn how dynamicsolo did it still.

Recognitions:
Gold Member
Staff Emeritus
 Quote by TbbZz I do understand that there must be three distinct zeroes. This tells me that the zeroes must all be real since every imaginary root (a+bi) means there is another imaginary root (a-bi). Because there is a double root, it would not be possible to have an imaginary root.
No, the must not be "three distinct zeros". You are told that there is a double root- there are only two distinct zeros!

 I worked for about 15 minutes on substituting R for the double root and r for the third root. However, I still can't seem to figure out how to solve for the missing coefficient. I know that: 2R + r = -C/4 <-- sum of roots 2Rr = 27/4 <-- product of roots But I'm not sure what to do next. P.S. HallsofIvy, I'm sure your method works, but unfortunately I don't understand the concept of it, so I'd like to learn how dynamicsolo did it still.
 Heh, I was counting the double root as two zeroes, since "the number of zeroes is equal to the degree of the polynomial." Still though, is there another formula I need to use other than -B/A and -1^n K/A?

Recognitions:
Homework Help
 Quote by TbbZz I do understand that there must be three distinct zeroes.
OK, we resolved this elsewhere.

 This tells me that the zeroes must all be real since every imaginary root (a+bi) means there is another imaginary root (a-bi). Because there is a double root, it would not be possible to have an imaginary root.
Just so. If one of the imaginary zeroes were repeated, one of those appearances would *not* have a "conjugate" partner, since there are only three zeroes in all.

 I know that: 2R + r = -C/4 <-- sum of roots
Careful: in the formula you are given, B is the coefficient of the *quadratic* term in the polynomial. That value is...?

 2Rr = 27/4 <-- product of roots
This is a *product* of the three roots. If one of them is repeated, that product would be...?

Recognitions:
Homework Help
 Quote by HallsofIvy Any cubic polynomial with a double (but not triple) root and leading coefficient 4, can be written 4(x-a)2(x-b)= 4x3-(8a+4b)x2+(4a2+8ab)x- a2b. Since you are told that the polynomial is 4x3+ cx- 27, you know that 8a+4b= 0, 4a2+ 8ab= c, and a2b= 27. That gives you three equations to solve for a, b, c.
This is certainly the other route to take. If the information given in the problem were somewhat different, this would be just as efficient as what I'm proposing. The fact that one zero is repeated here happens to make solving from the roots to the coefficients a bit easier; that wouldn't be the case in general, I suspect...

 dynamicsolo, I had to think for a minute to see why all the roots must be real!
It's sometimes worthwhile to use a bit of the "theory of functions" to get an idea of what the zeroes might look like and how complicated it will be to determine them.