an ambiguous quadratic equation question.

**quikwerk** · January 20th 2011, 12:10 PM

This question is very unclear as to what it wants for an answer, so I'd just like to check and see if my working is satisfactory.

The question:
Solve the equation $x^3+2x^2+x+2=0$ if -2 is a root.

All I did with my working is replace x with -2.
Is this going to satisfy the question?

**dwsmith** · January 20th 2011, 12:14 PM

Originally Posted by quikwerk

This question is very unclear as to what it wants for an answer, so I'd just like to check and see if my working is satisfactory.

The question:
Solve the equation $x^3+2x^2+x+2=0$ if -2 is a root.

All I did with my working is replace x with -2.
Is this going to satisfy the question?

Seems like a safe bet to me.

Also, this is a cubic not a quadratic.

If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.

**Quacky** · January 20th 2011, 12:58 PM

Originally Posted by dwsmith

Seems like a safe bet to me.

Also, this is a cubic not a quadratic.

If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.

I wouldn't even do that. Just equate coefficients.

If $x=-2$ is a root, then $(x+2)$ is a factor.

$x^3+2x^2+x+2=(x+2)(x^2 + ...)$

**quikwerk** · January 20th 2011, 01:04 PM

Originally Posted by dwsmith

Seems like a safe bet to me.

Also, this is a cubic not a quadratic.

If you are feeling ambitious, you can use synthetic division to take the equation down to a quadratic and find the remaining two roots.

I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?

**dwsmith** · January 20th 2011, 01:09 PM

Originally Posted by quikwerk

I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?

Yes, just show it is true when x = -2.

**Archie Meade** · January 20th 2011, 01:20 PM

Originally Posted by quikwerk

I'll probably find the other roots to be safe, but if I were to just go with my original working, all I would have to do is establish that the equation is true with -2 replacing x, yes?

No!!!

"Solve" the equation means

find all x for which the equation is true

hence, you must use the clue given to find the remaining solutions.

"Solve" means "find the solutions".

One solution is $x=-2$

$(x+2)\left(x^2+bx+c\right)=x^3+2x^2+x+2$

Clearly, this is zero if $x=-2$ since these are two ways to write the same expression.

If we multiply this out, $2c=2\Rightarrow\ c=1$

$bx^2+2x^2=2x^2\Rightarrow\ b=0$

$(x+2)\left(x^2+1\right)=x^3+x+2x^2+2$

The other roots are complex.

**quikwerk** · January 20th 2011, 02:10 PM

Originally Posted by Archie Meade

No!!!

"Solve" the equation means

find all x for which the equation is true

hence, you must use the clue given to find the remaining solutions.

"Solve" means "find the solutions".

One solution is $x=-2$

$(x+2)\left(x^2+bx+c\right)=x^3+2x^2+x+2$

Clearly, this is zero if $x=-2$ since these are two ways to write the same expression.

If we multiply this out, $2c=2\Rightarrow\ c=1$

$bx^2+2x^2=2x^2\Rightarrow\ b=0$

$(x+2)\left(x^2+1\right)=x^3+x+2x^2+2$

The other roots are complex.

My hopes are dashed!

Ok, so now I have to solve this cubic equation...
Would I be correct in supposing that one of the roots might be 2?

**Archie Meade** · January 20th 2011, 02:19 PM

Originally Posted by quikwerk

My hopes are dashed!

Ok, so now I have to solve this cubic equation...
Would I be correct in supposing that one of the roots might be 2?

No, one of the roots is $x=-2$

and it's the only real root.

There are also two complex roots, which are the solution to the 2nd factor being zero.

$x^3+2x^2+x+2=(x+2)\left(x^2+1\right)$

$x^3+2x^2+x+2=0\Rightarrow\ (x+2)\left(x^2+1\right)=0\Rightarrow\ x+2=0\;\;\;or\;\;\;x^2+1=0$

$(x+2)=0\Rightarrow\ x=-2$

If you are familiar with complex numbers, then you need to solve $x^2+1=0$

Otherwise, you need to be aware that $x=-2$ is the only real root

since there is no real solution to $x^2+1=0$

since $x^2=-1$ is the solution to that.

**quikwerk** · January 20th 2011, 02:45 PM

Originally Posted by Archie Meade

No, one of the roots is $x=-2$

and it's the only real root.

There are also two complex roots, which are the solution to the 2nd factor being zero.

$x^3+2x^2+x+2=(x+2)\left(x^2+1\right)$

$x^3+2x^2+x+2=0\Rightarrow\ (x+2)\left(x^2+1\right)=0\Rightarrow\ x+2=0\;\;\;or\;\;\;x^2+1=0$

$(x+2)=0\Rightarrow\ x=-2$

If you are familiar with complex numbers, then you need to solve $x^2+1=0$

Otherwise, you need to be aware that $x=-2$ is the only real root

since there is no real solution to $x^2+1=0$

since $x^2=-1$ is the solution to that.

Ok, I understand that, so what you are saying is that one root is "i" and another is something multiplied by i?

**Archie Meade** · January 20th 2011, 02:48 PM

Yes, and that "something" can only be one value.

What is $(-i)^2\;\;?$

Or

$(x+i)(x-i)=\;\;?$

**quikwerk** · January 20th 2011, 02:55 PM

Originally Posted by Archie Meade

Yes, and that "something" can only be one value.

What is $(-i)^2\;\;?$

Or

$(x+i)(x-i)=\;\;?$

$i*i=-1$
So you are saying that the solution set is:{-2, i, i^2}?

**Archie Meade** · January 20th 2011, 02:57 PM

Originally Posted by quikwerk

$i*i=-1$
So you are saying that the solution set is:{-2, i, i^2}?

Oh no!

Why do you reckon $i^2$ is a root ?

**Quacky** · January 20th 2011, 03:07 PM

Originally Posted by Archie Meade

Oh no!

Why do you reckon $i^2$ is a root ?

Lol, stop being a bully.

$i = \sqrt{-1}$
Therefore, $i^2 = -1$

What are the solutions to $\sqrt{4}$ ?

So, what are the solutions to $\sqrt{-1}$ ? One is $i$ . What is the other one?

**quikwerk** · January 20th 2011, 03:47 PM

Originally Posted by Quacky

Lol, stop being a bully.

$i = \sqrt{-1}$
Therefore, $i^2 = -1$

What are the solutions to $\sqrt{4}$ ?

So, what are the solutions to $\sqrt{-1}$ ? One is $i$ . What is the other one?

1*-1? I think?
Am I smart now?

**skeeter** · January 20th 2011, 03:59 PM

the roots of the equation are $x = -2$ , $x = i$ , and $x = -i$

imaginary roots form conjugate pairs.

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