Math problem, Algebra?

[dorcus]

Hello,

I am studying for examinations and encountered a problem I do not know how to solve:

Let w (omega) be a solution of the equation x^2 + x + 1 = 0

Then w^10 + w^5 + 3 = ?

 

I first replaced x with w (omega)

which yields

w^2 + w + 1 = 0

What I actually tried to do is solve the actual value of "w" using the quadratic equation. However, when I plug this into the 2nd equation, this yields "3", when the answer is 2.

 

Also, no calculators are allowed on the exam, so I can't be using the quadratic equation.

 

Any help is appreciated.

Thanks.

[D H]

Hint: What is x³ ?

[ydoaPs]

Hello,

I am studying for examinations and encountered a problem I do not know how to solve:

Let w (omega) be a solution of the equation x^2 + x + 1 = 0

Then w^10 + w^5 + 3 = ?

 

I first replaced x with w (omega)

which yields

w^2 + w + 1 = 0

What I actually tried to do is solve the actual value of "w" using the quadratic equation. However, when I plug this into the 2nd equation, this yields "3", when the answer is 2.

 

Also, no calculators are allowed on the exam, so I can't be using the quadratic equation.

 

Any help is appreciated.

Thanks.

 

Why would you replace x with w? I'd start by factoring the x equation, since we know that w is a solution.

[D H]

I wouldn't factor at all. That is the hard way to answer the question in post #2.

 

Another hint: Multiply both the left and right hand side of [math]x^2+x+1=0[/math] by [math]x-1[/math]. The right hand side is still zero. What is the left hand side?

[Bignose]

Also, no calculators are allowed on the exam, so I can't be using the quadratic equation.

 

I'm a little confused at this statement -- you don't need a calculator to use the quadratic equation, you just need the formula.

 

The solutions of [math]ax^2 + bx + c = 0[/math] are:

 

[math]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/math]

 

There is no need for a calculator, this is the solution. To show that, you should plug that into the equation and actually see for yourself that the expression becomes zero. You don't need a calculator to the algebraic manipulations.

[dorcus]

Hello,

I am sorry for not clarifying properly. What I meant that doing the quadratic equation would not be practical as taking the answer to the 10th power and adding it to the 5th power would require extensive calculation (The problems are designed to have "simple" calculations).

 

DH,

multiplying each side by x-1 would give:

 

x^3-1 = 0 or, x^3 = 1

which yields, x = cube root of 1, which is 1.

But I don't understand how that would work.

1^5 + 1^10 +3 = 5

 

When the answer is 2.

Am I doing something wrong?

Thanks.

[Malteaser]

I might be wrong, but doesn't the cube root of 1 have three answers - 2 of which are complex? That leads you back to the solution of the original quadratic, which is also complex.

 

The solution x=1 is the solution that we added when we multilied the original equation by x-1.

 

But I'm still stumped by the next move at the moment. I guess you could say that since

x^3 = 1 then you could divide 1 lot of x^3 from x^5 and 3 lots of x^3 from x^10.

 

This leaves the equation w^2 + w + 3 = ?

 

We know that x^2 + x = -1, from the original equation so substituting in we would end up with

 

-1 + 3 = ?

 

Which gives 2.

 

How's that?

[dorcus]

Hello,

That is slightly confusing.

So what you are suggesting is that I divide x^10 by x^3 3 times, while dividing x^5 one time. But are you allowed to do that?

I would have thought that you would need to divide x^10 and x^5 the same amount of times. Is there a property that perhaps I have not heard of that allows you to do that?

 

Thanks!

[D H]

multiplying each side by x-1 would give:

 

x^3-1 = 0 or, x^3 = 1

which yields, x = cube root of 1, which is 1.

x³=1 is correct. x=1 is not. There are three solutions to x³-1=0. x=1 is but one of them. The other two are the solutions to x²+x+1=0. You are looking for the solutions to the latter, not x=1. All three have one thing in common: Their cubes are one. Use this fact.

 

That is slightly confusing.

So what you are suggesting is that I divide x^10 by x^3 3 times, while dividing x^5 one time. But are you allowed to do that?

Multiplying or dividing any expression by one does not change the value of the expression one bit.

[Malteaser]

dorcus

 

I'm still not 100% sure we've taken the right approach myself - D H should be able to tell you that for sure.

 

As for your question. You are right I can't divide each part of a function by different expressions, that would be wrong and could get you in trouble.

 

What I can do is say that W^10 = W^3 * W^3 * W^3 * W.

 

But I know from D H's clever method of multilpying through by x-1 that x^3 and therefore W^3 is equal to 1. So all I need to do is substitute each W^3 with 1. I'm not dividing by anything.

 

What level Math is this by the way? Looks like UK A-level to me?

 

It might be worth trying to work out the answer using complex numbers to prove our answer - but I'm told you might need to use trig at some point and I can't remeber trig very well. D H - any help here?

[D H]

The hard way to show that the [math]\omega^3=1[/math] is to use the quadratic equation. The solutions to [math]w^2+w+1=0[/math] are

 

[math]w = \frac{-1 \pm \sqrt{1^2-4*1*1}}2

= -\,\frac 1 2 \pm \frac{\sqrt 3}2 i[/math]

 

Squaring yields

 

[math]w^2 = \frac 1 4 \mp 2\frac 1 2 \frac{\sqrt 3}2 i - \frac 3 4

= -\,\frac 1 2 \mp \frac{\sqrt 3}2 i[/math]

 

Multiplying by w again yields w³. An easier way is to recognize that w²=w^*: w squared is equal to the complex conjugate of w. Thus w³ = w*w^*, and this is just 1/4+3/4=1.

 

 

Writing a multiplicative factor of one or an additive factor of zero in a rather creative manner is a standard trick for solving apparently difficult problems. This is one of those problems.