A cubic equation and its roots

[mooncrater]

Homework Statement

The question says that :
Find the value of $a$ so that the equation $$x^3-6x^2+11x+a-6=0$$ has exactly three integer solitions.

Homework Equations

IF $p$,$q$,$r$ are the roots of this equation then:
$p+q+r=6$
$pq+pr+rq=11$
$pqr=6-a$

The Attempt at a Solution

I just don't know how to use these equations to find $a$.
So how to do this question?Are these equations of any use here?

 

[brainpushups]

I would simply factor it.

 

[mooncrater]

I would simply factor it.

But how can you simply factor it when there is an unknown $a$ in the equation?

 

[haruspex]

But how can you simply factor it when there is an unknown $a$ in the equation?

So factor the rest of it.

 

[Dick]

So factor the rest of it.

So that's pretty much just guessing that $a=0$, right? Or is there a better reason to do it? Suppose the constant were $a-7$?

 

[haruspex]

So that's pretty much just guessing that $a=0$, right? Or is there a better reason to do it? Suppose the constant were $a-7$?

It just struck me as something worth doing to obtain insight. It doesn't presume a=0, in fact it reveals an infinity of solutions. A minor modification solves a-7.

 

[ehild]

Find possible integer solutions which sum up to 6. Then select those for which pg+pr+qr=11.
You can eliminate one of the roots. Try completing squares.

 

[mooncrater]

Find possible integer solutions which sum up to 6. Then select those for which pg+pr+qr=11.
You can eliminate one of the roots. Try completing squares.

Thank you...Did what you said.
Leaving the zero values of $p,q,r$ , three possibilities were present:
$$p=1,q=1,r=4$$, $$p=1,q=2,r=3$$,$$p=q=r=2$$
Among which the second one satisfies the equation $$pq+qr+pr=11$$
So $pqr=6=6-a$
Therefore $a=0$.
Is this how you meant it to be done?

 

[ehild]

Thank you...Did what you said.
Leaving the zero values of $p,q,r$ , three possibilities were present:
$$p=1,q=1,r=4$$, $$p=1,q=2,r=3$$,$$p=q=r=2$$
Among which the second one satisfies the equation $$pq+qr+pr=11$$
So $pqr=6=6-a$
Therefore $a=0$.
Is this how you meant it to be done?

Yes, it was my first idea. There might be other solutions, including negative roots, but the problem text did not want you give or exclude more.

 

[haruspex]

Find possible integer solutions which sum up to 6. Then select those for which pg+pr+qr=11.
You can eliminate one of the roots. Try completing squares.

It says integer solutions, not natural numbers.
Factorising the terms other than a makes it obvious what all the solutions are.

 

[Dick]

It says integer solutions, not natural numbers.
Factorising the terms other than a makes it obvious what all the solutions are.

I still don't see it. You aren't just saying that $-a$ is the product of three consecutive integers, are you? That suffices to give one integer root. Doesn't say there are three unless $a=0$.

 

[haruspex]

I still don't see it. You aren't just saying that $-a$ is the product of three consecutive integers, are you? That suffices to give one integer root. Doesn't say there are three unless $a=0$.

Ah yes, not thinking straight.

 

[SammyS]

I have puzzled over how to use those equations for p, q, & r.

Use the factor theorem, plus perhaps Descartes's Sign Rule.

At most one negative root, that's if a > 6 . If -1 is a root, then a = 24.

If 0 is a root, a = 6. Look at the resulting quadratic.

If 1 is a root, a = __ , ...

 

[epenguin]

You could use psychology. Why are they giving you the equation in the form

$$x^3-6x^2+11x+a-6=0$$ .

?

The constant term is peculiar. It would be equivalent but normally more natural to ask what values of b give integral solutions of

$$x^3-6x^2+11x+b=0$$ .

It is because you might have to be a tad smarter to see how to do it when espressed that way. They are giving you a hint! I'll give you another. In the first equation the cubic without the a is quite easily factorised. Just knowing that it is makes it easier. Then if you think what a cubic looks like in a graph, you will see there is only a limited range of b within which it can have 3 real solutions and only a limited number of values for which it can have integer solutions.

 

[ehild]

The constant term is peculiar. It would be equivalent but normally more natural to ask what values of b give integral solutions of

$$x^3-6x^2+11x+b=0$$ .

It is because you might have to be a tad smarter to see how to do it when espressed that way. They are giving you a hint! I'll give you another. In the first equation the cubic without the b is quite easily factorised.

You meant factorising the part without a ?

 

[epenguin]

You meant factorising the part without a ?

Yes. Thank you, I have corrected that now.

 

[mooncrater]

You could use psychology. Why are they giving you the equation in the form?

The constant term is peculiar. It would be equivalent but normally more natural to ask what values of b give integral solutions of

$$x^3-6x^2+11x+b=0$$ .

It is because you might have to be a tad smarter to see how to do it when espressed that way. They are giving you a hint! I'll give you another. In the first equation the cubic without the a is quite easily factorised. Just knowing that it is makes it easier. Then if you think what a cubic looks like in a graph, you will see there is only a limited range of b within which it can have 3 real solutions and only a limited number of values for which it can have integer solutions.

Okay then , now I understand that this equation can also be solved without trial and error method. So I did it as:
$x^3-6x^2+11x-6=-a$
$x^3-5x^2-x^2+5x+6x-6=-a$
Which gives the equation:
$(x-1)(x-2)(x-3)+a=0$
Which has integer roots only if $a=0$
Is that what you want me to understand?

 

[epenguin]

Up to the last two lines. It has integer roots for a = 0 . The equation asks you find THE value of a that gives you that, so I suppose you have answered the question. You might worry a bit about whether this is the only possible value of a to give integer roots and how you would handle it if there were more than one value of a giving integer roots.

Perhaps there are cases like that, say for a cubic. Perhaps you could construct such cases. Or show that there can't be any. Depending on your time and interest. Polya principle of 'How to solve it' - when you have solved it, it's not finished. But at least try see if you can see in this case if 0 is the only value.

 

[mooncrater]

Up to the last two lines. It has integer roots for a = 0 . The equation asks you find THE value of a that gives you that, so I suppose you have answered the question. You might worry a bit about whether this is the only possible value of a to give integer roots and how you would handle it if there were more than one value of a giving integer roots.Perhaps there are cases like that, say for a cubic. Perhaps you could construct such cases. Or show that there can't be any. Depending on your time and interest. Polya principle of 'How to solve it' - when you have solved it, it's not finished. But at least try see if you can see in this case if 0 is the only value.

Okay I think I got what you want to say. We can have infinite values of $a$ to satisfy the equation. But since it's given that the equation has 3 roots therefore $a$ can have only one value that is $0$.
Is this what you want to say to me?

 

[epenguin]

Okay I think I got what you want to say. We can have infinite values of $a$ to satisfy the equation. But since it's given that the equation has 3 roots therefore $a$ can have only one value that is $0$.
Is this what you want to say to me?

No , I am saying it is not demonstrated and not all that obvious that there are no other values of a to give integral roots, and for someone wanting to develop a bit of mathematical muscle (of course you may consider you have better things to do with your time) it is of value to try and see further inside this.

 

[ehild]

See the plot of y = (x-1)(x-2)(x-3). It crosses the x-axis at 1, 2, 3, (the red dots), and the extrema are 0.385 and -0.385. Adding a number a it will shift up or down, or you can consider the x-axis shifting parallel. The coloured lines show the shifted axes.

There are less than 3 real roots if the x-axis is shifted too much...

If it is shifted up or down less than 0.385, two of the new real roots are between a pair of the red ones...

By the way, the constant term of the equation is the negative product of the roots. If you can add a which magnitude is less than 0.385, can be the constant term integer?

 

[sankalpmittal]

Homework Statement

The question says that :
Find the value of $a$ so that the equation $$x^3-6x^2+11x+a-6=0$$ has exactly three integer solitions.

Homework Equations

IF $p$,$q$,$r$ are the roots of this equation then:
$p+q+r=6$
$pq+pr+rq=11$
$pqr=6-a$

The Attempt at a Solution

I just don't know how to use these equations to find $a$.
So how to do this question?Are these equations of any use here?

Coming back after a long time. I would suggest you rather graph y=x^3-6x^2+11x-6 and y=-a on the paper. You can graph the cubic easily by checking at x=0 or may be y=0 and also checking maxima, minima , inflexion point etc. Then plot on the same paper y=-a. Check what values can you give a by seeing graph which would give exactly three integral solutions.