Question

Let \(a\) be a positive real number. If the following equation has a real root...

\(x^6 + 3a x^5 + (3a^2 + 3)x^4 + 6ax^3 + (3a^2 + 3)x^2 + 3ax + 1 + (ax)^3 = 0\)

Then find the minimum possible value of \(a\).

Answer 1

@Callisto There.

Answer 2

@perl

Answer 3

Hint: there's a pattern hidden in there. :)

Answer 4

any other usefufl hint ?

Answer 5

Do you see the pattern? You should be able to solve it if you do...

Answer 6

hmm i can only think of vieta formula and it looks hard

Answer 7

There's a thing with the coefficients. They repeat in a fascinating manner.

Answer 8

(x-a)^6=0

Answer 9

x=a

Answer 10

@Rishi123 I don't see how you reduced it to \((x-a)^6 =0\)

Answer 11

Let \(a\) be a positive real number.

Answer 12

Are you supposed to use grouping to solve for the polynomial?

Answer 13

Yes.

Answer 14

you can rewrite the lefthand side as
( x^2 + ax + 1)^3 = 0

Answer 15

You're damn right.

Answer 16

then use derivative to minimize that, i think

Answer 17

No need.

Answer 18

is that the pattern ? i wouldn't have figured that without wolfram :O

Answer 19

you can look at the discriminant ?

Answer 20

that has a real solution if
a^2 -4 (1)(1) > 0

Answer 21

>=

Answer 22

Pretty sure that perl has looked at Wolfram too, but my method was different.

Answer 23

Applied mathematicians unite! What sort of pure math madness is this? jk :)

Answer 24

so the minimum a, is 2?

Answer 25

God, I just lost all of my work...

Answer 26

Step 1: divide both sides by \(x^3\). You do this when you have mirroring coefficients.

Answer 27

\[\left(x^3 + \dfrac{1}{x^3}\right) + 3a \left( x^2 + \dfrac{1}{x^2}\right) + (3a^2 + 3) \left(x + \dfrac{1}{x}\right) + 6a + a^3 = 0\]

Answer 28

Let \(t = x + \dfrac{1}{x}\).

Then \(x^2 + \dfrac{1}{x^2} = t^2 - 2\).

And \(x^3 + \dfrac{1}{x^3}= t^3 - 3t\)

Answer 29

Plug those in and then with the cancellations and stuff, you'd be left with\[t^3 + 3at^2 + 3a^2 t + a^3 = 0\]\[\Rightarrow (t + a)^3 = 0\]

Answer 30

wow! nice thats pretty patterny xD

Answer 31

what was the minimum value for a?

Answer 32

\(x + \frac{1}{x} = -a \implies x^2 +ax+1 = 0 \)

\(D \ge 0\) yields \(a^2 - 4 \ge 0\) as desired

Answer 33

yes thats what I got too :)