Translation difficulties, Basic Integer Problem

**Zellator** · April 3rd 2011, 11:49 AM

Hi MHF!

I've been having some problems recently with word problems. I'm still not used to English, I guess.
So:

Consider the Integral expression in $x$ .
$P=x^3+x^2+ax+1$

Where $a=??$ and the value of $P$ is a rational number for any $x$ which satisfies the equation $x^2+2x-2=0$ , and in this case the value of $P$ is:

I'm sorry if there's any typos in there, I've copied this some time ago.

What "the integral expression in $x$ " means?

$\int[x^3+x^2+ax+1]$ or maybe $d/dx [x^3+x^2+ax+1]$ ?

Not having a clue on how to start, I've tried to resolve $x^2+2x-2=0$ .
Still with $-1+(3)^(1/2)$ and $-1-(3)^(1/2)$ , no luck.

$P$ represents many numbers, right? But what number does it actually ask for?
Is $P=-1+(3)^(1/2)$ and $-1-(3)^(1/2)$ ?

$(3)^(1/2)$ is supposed to be a root

*completely lost*

Thanks :3

**Sambit** · April 3rd 2011, 10:58 PM

Originally Posted by Zellator

What "the integral expression in $x$ " means?

$\int[x^3+x^2+ax+1]$ or maybe $d/dx [x^3+x^2+ax+1]$ ?

The first one is integration (the second one is differentiation; not integration)

Originally Posted by Zellator

$P$ represents many numbers, right? But what number does it actually ask for?
Is $P=-1+(3)^(1/2)$ and $-1-(3)^(1/2)$ ?

The roots of the quadratic equation are $\frac{-2\pm\sqrt{12}}{2}$

**Opalg** · April 4th 2011, 12:04 AM

Originally Posted by Zellator

Hi MHF!

I've been having some problems recently with word problems. I'm still not used to English, I guess.
So:

Consider the Integral expression in $x$ .
$P=x^3+x^2+ax+1$

Where $a=??$ and the value of $P$ is a rational number for any $x$ which satisfies the equation $x^2+2x-2=0$ , and in this case the value of $P$ is:

I'm sorry if there's any typos in there, I've copied this some time ago.

What "the integral expression in $x$ " means?

The word "integral" has two completely different meanings in English. It can mean something to do with integration, or it can mean something to do with integers.

My guess is that this is an algebra problem, not a calculus problem, and that "the integral expression in $x$ " means "a function of $x$ whose value is an integer". So forget about integration, and think about integers. As you correctly say, the solutions of the equation $x^2+2x-2=0$ are $x = -1\pm\sqrt3$ . Plug those values into the expression $P$ , and find the value of $a$ for which the answer is an integer. In other words, choose $a$ so that terms involving $\sqrt3$ cancel out and you are left with an "integral" answer.

**Zellator** · May 27th 2011, 04:47 AM

I'll have to revive this thread since I'm still having difficulties.

Opalg suggested me to plug the values $x = -1\pm\sqrt3$ in the formula.
I'm kind of confused, since the question says this is a value of P and not of x

So what would be right?
$x = -1\pm\sqrt3=x^3+x^2+ax+1$

or probably
$P=(-1\pm\sqrt3)^3+(-1\pm\sqrt3)^2+a(-1\pm\sqrt3)+1$
In this case, what would happen to P?
Would it equal zero? Would it equal $-1\pm\sqrt3$ ?

Both cases, I still get a racional number that is nowhere near $a=-4$

This sqrt3 is really annoying.
Thanks!

**HallsofIvy** · May 27th 2011, 05:36 AM

Originally Posted by Zellator

I'll have to revive this thread since I'm still having difficulties.

Opalg suggested me to plug the values $x = -1\pm\sqrt3$ in the formula.
I'm kind of confused, since the question says this is a value of P and not of x

So what would be right?
$x = -1\pm\sqrt3=x^3+x^2+ax+1$

or probably
$P=(-1\pm\sqrt3)^3+(-1\pm\sqrt3)^2+a(-1\pm\sqrt3)+1$

This second one.

In this case, what would happen to P?
Would it equal zero? Would it equal $-1\pm\sqrt3$ ?

It must be equal to an integer. It might be 0 but certainly not $-1\pm\sqrt{3}$

Both cases, I still get a racional number that is nowhere near $a=-4$

??? No one has said anything about a being equal to -4.

This sqrt3 is really annoying.
Thanks!

If $x= -1+ \sqrt{3}$ what is $x^3$ ? What is $x^2$ ? What is $x^3+ x^2+ ax+ 1$ ? What value of a will make that an integer?

**Zellator** · May 27th 2011, 07:50 AM

Hi HallsofIvy!
hahah the a=-4 part is the answer!
Sorry If I haven't made myself clear.

Originally Posted by HallsofIvy

If $x= -1+ \sqrt{3}$ what is $x^3$ ? What is $x^2$ ? What is $x^3+ x^2+ ax+ 1$ ? What value of a will make that an integer?

Right! $x^3$ = $-1+2\sqrt{3}-3 +\sqrt{3}-6+3\sqrt{3}$ when $x= -1+ \sqrt{3}$
$x^2$ = $1-2\sqrt{3}+3$

I'm not sure where you want to get here, sorry.
I'm still not sure about the end of that equation.
It also says that a is a rational number.
This has been omited so far, I'm sorry.

Working on the question we get.
$2\sqrt{3}-5+a(1+\sqrt{3})=??$

Ok, so if we equal this to zero and square both sides we can get
a^2=73/4
This gives something close to 4.23...
Close enough, but is this conclusive?
We still have to find the value of P when a is in our desired value.

Edit: Can some moderator please move this topic to it's right place?
This looks like but is not a calculus problem.

**HallsofIvy** · May 27th 2011, 08:53 AM

Originally Posted by Zellator

Hi HallsofIvy!
hahah the a=-4 part is the answer!
Sorry If I haven't made myself clear.

Right! $x^3$ = $-1+2\sqrt{3}-3 +\sqrt{3}-6+3\sqrt{3}$ when $x= -1+ \sqrt{3}$
$x^2$ = $1-2\sqrt{3}+3$

Okay- $x^2= 4- 2\sqrt{3}$ but that is not what I get for $x^3$ . For $x^3$ you should have $(-1+ \sqrt{3})(4- 2\sqrt{3})= -4+ 4\sqrt{3}+ 2\sqrt{3}- 6$ , not what you have.

I'm not sure where you want to get here, sorry.
I'm still not sure about the end of that equation.
It also says that a is a rational number.
This has been omited so far, I'm sorry.

Working on the question we get.
$2\sqrt{3}-5+a(1+\sqrt{3})=??$

Ok, so if we equal this to zero and square both sides we can get
a^2=73/4
This gives something close to 4.23...
Close enough, but is this conclusive?
We still have to find the value of P when a is in our desired value.

No, it does not equal 0- only the coefficient of $\sqrt{3}$ has to be 0 in order that P be an integer.

Edit: Can some moderator please move this topic to it's right place?
This looks like but is not a calculus problem.

**Zellator** · May 27th 2011, 10:01 AM

Originally Posted by HallsofIvy

Okay- $x^2= 4- 2\sqrt{3}$ but that is not what I get for $x^3$ . For $x^3$ you should have $(-1+ \sqrt{3})(4- 2\sqrt{3})= -4+ 4\sqrt{3}+ 2\sqrt{3}- 6$ , not what you have.

It is, actually!
I left it this way because it is easier to make it disappear in the main question, this is about taste hahah
$-1-3-6+2\sqrt{3}+\sqrt{3}+3\sqrt{3}=-4+6\sqrt{3}-6$

Later on you have
$4-2\sqrt{3}$

Which gets to the final
$4\sqrt{3}-5+a(-1+\sqrt{3})$

not 2\sqrt{3} as I have mentioned earlier.

No, it does not equal 0- only the coefficient of $\sqrt{3}$ has to be 0 in order that P be an integer.