Solve the roots of a cubic polynomial?

Question

I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:

"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

score 2 · Accepted Answer · answered May 15 '13 at 10:20

2

It means the roots are of the form $a, ar, ar^2$.

Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients $$ 14 = 2 + 4 + 8, \qquad 56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8. $$

answered May 15 '13 at 10:20

Andreas Caranti

68,873

Thanks, but are there different types of geometric progessions? – missiledragon May 15 '13 at 10:21
@missiledragon, not that I am aware of, it means a sequence of the form $a_{n} = a r^{n}$ for some $a, r$, see the Wikipedia link. – Andreas Caranti May 15 '13 at 10:23
Thanks, if you don't mind, could you briefly explain what an arithmetic progression is? I had a similar question with it. – missiledragon May 15 '13 at 10:25
@missiledragon, the Wikipedia is your friend http://en.wikipedia.org/wiki/Arithmetic_progression it means a sequence of the form $a_{n} = a + n r$, for fixed $a, r$, and $n \ge 0$. – Andreas Caranti May 15 '13 at 10:27
Alright, thanks for your help – missiledragon May 15 '13 at 10:32
@missiledragon, you're welcome. – Andreas Caranti May 15 '13 at 10:32

lab bhattacharjee · Answer 2 · 2013-05-15T11:30:47.637

1

Let the roots be $a,a\cdot r,a\cdot r^2$

Using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$

and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$

On division, $ar=4$ as $a\cdot r\ne0$

$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{ or }\frac12$$

Alternatively, using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$

So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$

Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation

$\implies a\cdot r=4\iff a=\frac4r$

Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$

edited May 15 '13 at 11:30

answered May 15 '13 at 10:21

lab bhattacharjee

274,582

Why is $a$ multiplied by $r$ ? – missiledragon May 15 '13 at 10:23
@missiledragon, http://mathworld.wolfram.com/GeometricSeries.html – lab bhattacharjee May 15 '13 at 10:29
@missiledragon, http://en.wikipedia.org/wiki/Geometric_progression – lab bhattacharjee May 15 '13 at 10:30
Wait, isn't it $ar(1+r+r^2)=56$ supposed to have $ar^2$ instead? – missiledragon May 15 '13 at 11:26
Yep, I used his formula. – missiledragon May 15 '13 at 11:31
No worries. Are you going to fix the rest of the answer? – missiledragon May 15 '13 at 11:32
How did you get: $4(1+r+r^2)/r$? – missiledragon May 15 '13 at 11:35
let us continue this discussion in chat – lab bhattacharjee May 15 '13 at 11:36

Solve the roots of a cubic polynomial?

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