Complex numbers: if 1+i is root of f(x) = ax^3+x^2+2x+b, find values of a, b

[lukas.k]

Hello,
got stuck on this problem. I would appreciate some help.

f(x) = ax^3 +x^2+2x+b

given that 1+i is a root of the equation f(x) = 0.​

find the value of a and b ?​

I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root)

not really sure what to do next.

can anyone help please ?

 

[stapel]

Hello,
got stuck on this problem. I would appreciate some help.

f(x) = ax^3 +x^2+2x+b

given that 1+i is a root of the equation f(x) = 0.​

find the value of a and b ?​

I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root)

not really sure what to do next.

can anyone help please ?

Recall that you get complex-valued solutions when the Quadratic Formula here had a negative inside the square root. This means, in particular, that the Quadratic Formula was applied. And you know that the Quadratic Formula spits out plus/minus versions of the roots. So, if x = a - bi is a root, then so also is x = a + bi. So find the second root which is based on the first one they gave you.

Recall the definition of a "root". In particular, it is a zero. So what is the value of f(x) when you plug in either of the two known roots?

Plugging this value into the functional statement, along with the roots, will give you two equations in two unknowns. Solve for "a" and "b".

If you get stuck, please reply showing your steps in following the above instructions. Thank you!

 

[pka]

Hello,
got stuck on this problem. I would appreciate some help.
f(x) = ax^3 +x^2+2x+b

given that 1+i is a root of the equation f(x) = 0.​

find the value of a and b ?​

I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root)

What you have done assumes that a & b are real numbers (assuming conjugate root)
Are you given that fact?

 

[Steven G]

Hello,
got stuck on this problem. I would appreciate some help.

f(x) = ax^3 +x^2+2x+b

given that 1+i is a root of the equation f(x) = 0.​

find the value of a and b ?​

I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root)

not really sure what to do next.

can anyone help please ?

I think that your problem is writing *(real root). Rather write *(x-real root) or (x-c), where c is a real number.

So just do the multiplication! Just make sure that you pick c so that the coefficient of x^2 is 1

 

[pka]

I think that your problem is writing *(real root). Rather write *(x-real root) or (x-c), where c is a real number. So just do the multiplication! Just make sure that you pick c so that the coefficient of x^2 is 1

But to my point. The conjugate root theorem only applies to polynomials over the real field. Unless we are given that a & b are real numbers, that theorem cannot be used.

 

[Ishuda]

Hello,
got stuck on this problem. I would appreciate some help.

f(x) = ax^3 +x^2+2x+b

given that 1+i is a root of the equation f(x) = 0.​

find the value of a and b ?​

I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root)

not really sure what to do next.

can anyone help please ?

What you have said [that the roots are 1+i, 1-i, and a real root] imply that both a and b are real. So
a (1+i)³ + (1+i)² + 2 (1+i) + b = 0
Now
(1+i)² = 2i
(1+i)³ = -2 + 2i
thus
a (1+i)³ + (1+i)² + 2 (1+i) + b = a (-2 + 2i) + 2i + 2 + 2 i + b = (b - 2 a + 2) + 2 (a + 1) i = 0

Can you continue from there?

EDIT: red is mistake, should be a+2

 

[pka]

What you have said [that the roots are 1+i, 1-i, and a real root] imply that both a and b are real.

Once again, the given is only that (1+i) is a root. There is no reason to assume that a or b is not complex. Therefore, it incorrect to assume that (1-i) is also a root.

 

[Steven G]

But to my point. The conjugate root theorem only applies to polynomials over the real field. Unless we are given that a & b are real numbers, that theorem cannot be used.

Prof, yes you are certainly correct. Thank you!

 

[pka]

pka,
I believe you are incorrect in your assertion. I think there is every reason to know that a and b are real valued.

Let \(\displaystyle \large a=1~\&~b=-6\bf{i}\)
Look at the solution.

 

[Steven G]

I think somewhere in the problem or the section it stated that a and b are real numbers. But the way it was posted it did not mention it. I was wrong for assuming it (2 minutes in the penalty box for me as I don't like the corner) and pka is 100% correct.

There really is no point to argue over this. The post involved complex numbers--so complex numbers are known to the OP-and it did not say that a and b are reals so there is no reason to assume complex roots occurs in pairs.

 

[Ishuda]

Let \(\displaystyle \large a=1~\&~b=-6\bf{i}\)
Look at the solution.

As I understand it we have
GIVEN:
"... I know that ax^3+x^2+2x+b = (z-(1+i))*(z-(1-i))*(real root)
so ax^3+x^2+2x+b =(z^2 -2z+2)*(real root) ..."
Your example does not have the two given roots of 1+i and 1-i, see.
http://www.wolframalpha.com/input/?i=x^3+++x^2+++2+x+++6i+=+0

So starting at
f(x) = a (x² - 2x + 2) (x-c) = a [x³ - (2+c) x² + 2 (1 + c) x - 2 c]
we have
-a (2+c) = 1 from coefficient of x²
a (1+c) = 1 from coefficient of x
So
-2-c = 1+c
and
c = -3/2
a = -2
From constant coefficient
b = -2 a c = -6
See
http://www.wolframalpha.com/input/?i=-2+x^3+++x^2+++2+x+-+6+=+0


NOTE: It seems obvious to me that the (real root) from the OP means 'a (x-c) where a and c are real' but even if we don't (initially) allow that a is real, the statement was the other root was real so at least c is real. However even that is not needed as shown above.