# Homework Help: Solve for x

1. Jan 7, 2016

### emrys

1. The problem statement, all variables and given/known data
(x2+3x+3)1/3 + (2x2+3x+2)1/3 = 6x2+12x+8
2.Relevant equations

3. The attempt at a solution
(x2+3x+3)1/3 -1 +(2x2+3x+2)1/3 -1 = 6x2+12x+6
(x2+3x+2)/((((x2+3x+3)1/3)2 + (x2+3x+3)1/3 +1) + (2x2 +3x+1)/((((2x2+3x+2)1/3)2)+(2x2+3x+2)1/3 +1) -6(x+1)2=0
then
x=-1
or
(x+2)/((((2x2+3x+2)1/3)2)+(2x2+3x+2)1/3 +1)-6(x+1)=0 (*)
now there is 1 root left it is also -1 but no common factor in (*)
sorry i am not good at english

2. Jan 7, 2016

### BvU

Hello emrys,

To be honest, I have no idea how to deal with this kind of equation analytically, and I can't follow your first step (*), so I simply plotted the terms and I see that only -1 is a solution.

Multiplicity is 2, also from visual inspection.

Is there a question in your post ?

(*) perhaps you can explain, then I will learn something from this too !

Last edited: Jan 7, 2016
3. Jan 7, 2016

### emrys

for
A= (x2+3x+2)1/3
B= (2x2+3x+2)1/3
then thanks to calculator i know that x = -1
so i substitute x for -1 to find value of A,B then subtract exactly that value to find common factor (x+1):
A -1 + B -1 = 6x2 +12x+6
(A-1)(A2+A*1+12)/(A2+A*1+12) + (B - 1)(B2+B*1+12)/(B2+B*1+12)-6(x+1)2=0
(A3-13)/(A2+A*1+12) + (B3-13)/(B2+B*1+12) -6(x+1)2=0
then you got
x=-1
or
(x+2)/(A2+A*1+12) + (2x+1)/(B2+B*1+12) - 6(x+1)=0(*)

4. Jan 7, 2016

### BvU

A= (x2+3x+3)1/3

So basically you are also solving this numerically ? Finding x = -1 to satisfy the original equation ?
Then what is the purpose of the remainder of the work ?

5. Jan 7, 2016

### emrys

solve(*) to find the final root (also = -1).
to end the problem you must find all the roots or the equation becomes an impossible equation.
and there is 1 root left in (*).
In brief, i want you guys to help me solve (*) (normally do it like you didnt know about x=-1)

6. Jan 7, 2016

### Isaac0427

I am actually around this level of math. My first approach at this question would be to graph it and get the solutions by treating it like a system of equations, so $y=(x^2 + 3x + 3)^1/3 + (2x^2 + 3x + 2)^1/3$ and $y = 6x^2 + 12x + 8$ (sorry if I did the latex wrong). If you need to do it algebraicly, it still may help having the answer to help guide you in solving it.

7. Jan 7, 2016

### Isaac0427

And I did do the latex wrong. It should be to the 1/3 power not to the 1st power decided by 3.

8. Jan 7, 2016

### Ray Vickson

I have a very strong suspicion that your equation cannot be solved "analytically" in any easy way, so if I were trying to do it the very first thing I would do is plot the functions. That reveals x = -1 as a double root (a fact that can be verified analytically, once we know what we are looking for). That is how I would normally do it if I did not know about x = -1 ahead of time.

9. Jan 9, 2016

### emrys

Use cauchy inequality, we can solve this easily.

10. Jan 9, 2016

### Ray Vickson

How can you use an inequality to solve an equation? You should at least show how you would do it.

11. Jan 9, 2016

### SammyS

Staff Emeritus
@Isaac0427
Regarding the LaTeX:

To include the /3 in the the exponent, place { } around the 1/3 like so: ^{1/3} Doing that to your first LaTeX expression gives:

$y=(x^2 + 3x + 3)^{1/3} + (2x^2 + 3x + 2)^{1/3}\$​

12. Jan 10, 2016

### emrys

we will do it again from beginning
we have
(x2+3x+3)1/3>0 ∀x∈ℝ
(2x2+3x+2)1/3>0 ∀x∈ℝ
and 6x2+12x+8>0 ∀x∈ℝ
then
apply cauchy inequality (A+A1+A2+...+An)/n ≥(A*A1*A2*....*An)1/n; (A,A1,A2,...,An≥0)
6x2+12x+8=(1*1(x2+3x+3))1/3+(1*1(2x2+3x+2))1/3≤(x2+3x+3+1+1)/3+(2x2+3x+2+1+1)/3=x2+2x+3
⇔6x2+12x+8≤x2+2x+3
⇔52+10x+5≤0
⇔5(x+1)2≤0
⇔x=-1

13. Jan 10, 2016

### SammyS

Staff Emeritus
Looks good.

There is a typo in the indicated line. Of course, it should read:

5x2+10x+5≤0​

14. Jan 10, 2016

### emrys

Your method works accidentally in this one special case. In most cases it will not work. For instance, if you change the equation to $A + B = 6x^2+12x + 6$ (where $A,B$ are as before) you will find that $x^2 + 2 x + 3 \geq 6 x^2 + 12 x + 6$ at any solution $x$ of the original equation. This is true, but it does not help you to find $x$. Now we need to use a numerical method, to find there are two roots: $x = -1.6116857121542768293$ and $x = -0.37972871671526324113$.