Complex number as a root and inequality question

[Darken1]

Question 1:
(a) Show that the complex number i is a root of the equation
x^4 - 5x^3 + 7x^2 - 5x + 6 = 0
(b) Find the other roots of this equation

Work:
Well, I thought about factoring the equation into (x^2 + ...) (x^2+...) but I couldn't do it. Is there a method for that? Anyways the reason I wanted to do that was to use the quadratic formula afterwards where I guess one of the zeroes is x=i.Edit: Think I solved my own question 2 upon writing this
Question 2:
Let f(x) = (4-x^2)/(4-x^(1/2))

(a) State the largest possible domain for f.

(b )Solve the inequality f(x) greater than or equal to 1

Work:
Domain: [0,16) U (16,infinity)
x^(1/2) so x can't be negative (greater than or equal to 0) and it's a fraction so the denominator cannot equal to 0.

Multiplied denominator to the other side
4-x^2 >= 4-x^(1/2)
Eventually arrived to
x^2-x^(1/2) <=0 and 0<=x^2+x^(1/2)
Took the first equation and changed the first term
(x^(1/2))^4-x^(1/2)<=0
x<=1

 

[Joppy]

Question 1:
(a) Show that the complex number i is a root of the equation
x^4 - 5x^3 + 7x^2 - 5x + 6 = 0

I think we can cheat a bit here and use the question itself to help us! If \(\displaystyle i\) is a root of the equation, then there must be a factor of \(\displaystyle (x^2 + 1)\). Does this make sense? In other words, we want to factor the expression, and thus determine the roots of the equation. One of which, will be i.

I get, \(\displaystyle (x^2 + 1)(x - 2)(x - 3) = 0\) which has real roots at 2, and 3, and non-real roots at \(\displaystyle \pm i\) (the advantage of this one is that we solve b) also).

Alternatively, and as you have said, we can simply let \(\displaystyle x = i\), and see if whether we come across any strange results.

\(\displaystyle i^4 - 5i^3 + 7i^2 - 5i + 6 = 0\)

\(\displaystyle i^2i^2 - 5i^2i + 7i^2 -5i + 6 = 0\)

\(\displaystyle 1 +5i - 7 - 5i + 6 = 0\)

\(\displaystyle 0 = 0\) and hence, \(\displaystyle i\) is indeed a root of the equation.There are a few different methods for solving these, which ones have you learned for dealing with these polynomials?

 

[Darken1]

I think we can cheat a bit here and use the question itself to help us! If \(\displaystyle i\) is a root of the equation, then there must be a factor of \(\displaystyle (x^2 + 1)\). Does this make sense? In other words, we want to factor the expression, and thus determine the roots of the equation. One of which, will be i.

I get, \(\displaystyle (x^2 + 1)(x - 2)(x - 3) = 0\) which has real roots at 2, and 3, and non-real roots at \(\displaystyle \pm i\).

There are a few different methods for solving these, which ones have you learned for dealing with these polynomials?

For x^4+... I just try to convert it to (x^2+...)(x^2+...) format by guessing tbh. I know how to factor cubes, the quadratic formula, and how to divide one polynomial by another.

 

[Joppy]

For x^4+... I just try to convert it to (x^2+...)(x^2+...) format by guessing tbh. I know how to factor cubes, the quadratic formula, and how to divide one polynomial by another.

That's a good start, and yes, a bit of educated guesswork can come in handy at times.

I'll run through how i did it, and then perhaps someone else can provide a more formal or general method for solving. A few main things i used to help me solve were,

1. \(\displaystyle (x^2 + 1)\) must be a factor (given in the question)
2. The product of all the factored constants must equal 6.

Because of 1., i assumed that there must be at least two other factors of the form \(\displaystyle (x+b)(x+a)\). I did so, because we know there can't be any more x terms right? Otherwise we would have an \(\displaystyle x\) to the power of a number greater than 4, which isn't in our original expression. So we have,

\(\displaystyle (x^2 + 1)(x+a)(x+b) = 0\)

At this point i got lazy and just 'trialed and errored' a bit, taking a and b as factors of 6 (2.). and knowing that the sum of \(\displaystyle a\) and \(\displaystyle b\) must equal \(\displaystyle -5\). i.e., \(\displaystyle a + b = -5\) and \(\displaystyle ab = 6\), \(\displaystyle -2\) and \(\displaystyle -3\) are the obvious candidates here :p.

This approach is pretty adhoc, and i wouldn't recommend it (i'll probably get in trouble for it :p) while you're still learning. Sparkfun offers some not bad advice on factoring polynomials of degree 3 and 4.

 

[Darken1]

That's a good start, and yes, a bit of educated guesswork can come in handy at times.

I'll run through how i did it, and then perhaps someone else can provide a more formal or general method for solving. A few main things i used to help me solve were,

1. \(\displaystyle (x^2 + 1)\) must be a factor (given in the question)
2. The product of all the factored constants must equal 6.

Because of 1., i assumed that there must be at least two other factors of the form \(\displaystyle (x+b)(x+a)\). I did so, because we know there can't be any more x terms right? Otherwise we would have an \(\displaystyle x\) to the power of a number greater than 4, which isn't in our original expression. So we have,

\(\displaystyle (x^2 + 1)(x+a)(x+b) = 0\)

At this point i got lazy and just 'trialed and errored' a bit, taking a and b as factors of 6 (2.). and knowing that the sum of \(\displaystyle a\) and \(\displaystyle b\) must equal \(\displaystyle -5\). i.e., \(\displaystyle a + b = -5\) and \(\displaystyle ab = 6\), \(\displaystyle -2\) and \(\displaystyle -3\) are the obvious candidates here :p.

This approach is pretty adhoc, and i wouldn't recommend it (i'll probably get in trouble for it :p) while you're still learning. Sparkfun offers some not bad advice on factoring polynomials of degree 3 and 4.

I can't see the url outside of me replying. Anyways what about dividing the original equation by (x^2 + 1) and then attempting to factor whatever it is?

 

[Joppy]

I can't see the url outside of me replying.

Not sure about that. Maybe i didn't embed them correctly.

Anyways what about dividing the original equation by (x^2 + 1) and then attempting to factor whatever it is?

Yes long division is a pretty common method. It's quite robust too.