1. Jul 19, 2016

### sooyong94

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

2. Relevant equations

3. The attempt at a solution
I did use Vieta's formula, and got a set of equations in terms of α, however the tricky bit is to deal with the cube root. Then I tried the quadratic formula, but the algebra becomes too tricky (due to nested square roots).

2. Jul 19, 2016

### haruspex

3. Jul 19, 2016

### sooyong94

4. Jul 19, 2016

### haruspex

I suggest either of two approaches:
1. Use one of the Vieta equations to eliminate alpha from the other, so you have an equation with no occurrences of alpha. Yes, it will be messy, but it should lead to the target.
2. Rewrite the target equation to consist of combinations of b/a and c/a and see if you can simplify it to a tautology. Reversing those steps should then be a proof.

5. Jul 19, 2016

### Delta²

$$\frac{-b^3}{a^3}=(x+\sqrt{x})^3=x^3+x^{3/2}+3(x^2+x^2\sqrt{x})=\frac{c^2}{a^2}+\frac{c}{a}+3(x^2+x\frac{c}{a})$$

I am also stuck here, maybe replace $x^2=-\frac{b}{a}x-\frac{c}{a}$...
or $x=(\frac{c^2}{a^2})^{\frac{1}{3}}$
Well maybe you can find the continuation, I am out of ideas here.

6. Jul 19, 2016

### Delta²

Now that I check again it is enough to prove that $x^2+x\frac{c}{a}=-\frac{bc}{a^2}$ I wonder how we can prove that...

Last edited: Jul 19, 2016
7. Jul 19, 2016

### SammyS

Staff Emeritus
Use your results from Vieta's formula.

Then cubing $\ \alpha+\sqrt \alpha \$ should help greatly. Keep factors of $\ \alpha\sqrt \alpha \$ wherever they occur.

$\displaystyle \left(\alpha+\sqrt \alpha \right)^3 = \alpha^3+3\alpha( \alpha\sqrt \alpha ) +3\sqrt \alpha ( \alpha\sqrt \alpha )+\alpha\sqrt \alpha \ .$

From there it's pretty straight forward.

8. Jul 19, 2016

### Delta²

Ah I was so close too but SammyS got me.

$x^2+x\frac{c}{a}=x\sqrt{x}\sqrt{x}+x\frac{c}{a}=\frac{c}{a}(\sqrt x+x)$

9. Jul 20, 2016

### haruspex

Just to illustrate the approaches I mentioned in post #4, not because they're any quicker but because they are goal-directed, not requiring any inspired insight:

1. Eliminate alpha using Vieta:
$\alpha=(\frac ca) ^{\frac 23}$
$(\frac ca) ^{\frac 23}+(\frac ca) ^{\frac 13}=-\frac ba$
Cubing
$(\frac ca)^2+3(\frac ca) ^{\frac 53}+3(\frac ca) ^{\frac 43}+\frac ca=-(\frac ba)^3$
$=(\frac ca)^2+3(\frac ca)((\frac ca) ^{\frac 23}+(\frac ca) ^{\frac 13})+\frac ca=(\frac ca)^2+3(\frac ca)(\alpha+\sqrt {\alpha})+\frac ca$
Etc.

2. Reverse engineering:
$b^3=ac(3b-a-c)$
$(\frac ba)^3=\frac ca(3\frac ba-1-\frac ca)$
Substitute throughout using alpha, simplify to obtain a tautology, then observe that the same steps can be run in reverse.

10. Jul 20, 2016

### Delta²

yes well basically your approach 1. is what we all 3 did, SammyS, me and you.

Approach 2 is quite interesting, I wouldn't have my mind thinking like that at first glance.

11. Jul 20, 2016

### Ray Vickson

Write the initial quadratic as $x^2 + b' x + c' = 0$, where $b' = b/a$ and $c' = c/a$. The condition to be verified is $b'^3 = c'(3 b' - 1 - c')$. (This is just the quadratic with leading coefficient = 1, etc.

OK, so the quadratic is $p(x) = (x-v)(x-v^2)$ for some $v$; that means that $b' = -v - v^2$ and $c' = v^3$. The rest follows by straightforward algebra.

Last edited: Jul 20, 2016
12. Jul 20, 2016

### epenguin

Maybe (tex typo?) you mean just x2 + b' x + c' = 0$, where$b' = b/a$and$ c' = c/a. ?

OK I have done it - easier than following anyone else's argument which I find like someone else's unmade bed! But equivalent I think.

Moral of the story, see thread title, avoid radicals when you can.

Secondly, almost always for algebraic equations make the leading coefficient 1, divide by a, work on the equation e.g.
x2 + b'x + c' = 0. You will get simpler intermediate expressions, more likely to be related to textbook statements, and since the roots are still the same, what is true of a, b,c is true of 1, b', c' . So you can easily recognise whether you are getting towards the right results.

13. Jul 20, 2016

### sooyong94

Thanks - I managed to work this out easily. ;)

14. Jul 20, 2016

### SammyS

Staff Emeritus
That's good.

What did you get for the second part? What did you get for y, and what did you get for roots of the resulting equation(s) ?

15. Jul 20, 2016

### sooyong94

I got y=-1 and y=6, and the resulting roots are
x=1/9 and x=-1/3

x=4/9 and x=-2/3

16. Jul 21, 2016

### SammyS

Staff Emeritus
Those are indeed the solutions that result, and it is true that $\displaystyle \ \left(-\frac 13 \right)^2=\frac 19 \,,\$ it's also the case that strictly speaking, $\displaystyle \ \sqrt \frac 19 \ne -\frac 13 \ .$

Similar results hold for the roots resulting from the other y solution.

Last edited: Jul 21, 2016
17. Jul 21, 2016

### SammyS

Staff Emeritus
Now, let's suppose that we are to be strict about using the principle square root here, and that we have written the quadratic equation with b' and c' as suggested by Ray Vickson and epenguin .
That is to say, the quadratic equation $\ x^2 + b\,' x + c' = 0 \$ has solutions, $\ \alpha \$.and $\ \sqrt \alpha \ .$

Since $\ \alpha\$.and $\ \sqrt \alpha\$ are both positive, b' must be negative, since $\ -\alpha-\sqrt \alpha\ = {b}\,' \ .$ Similarly, c' must be positive.

For the equation, $\ 27x^2+6x-(y+2)=0\$ and the values that sooyong94 got for y, we have positive b' and negative c' .

Using $\ -6\$ for b, so we have $\ 27x^2-6x-(y+2)=0\$, will give two irrational values for y, one of which does give two positive (and irrational) values for x.

To get rational solutions for x, use $\ b=-12\$ or $\ b=-30\ .$