If $a$ and $b$ are the roots of the equation $x^2-x+1=0$, then $a^{2009}+b^{2009}$ is
One way to solve this is to multiply the equation by $x+1$ and obtain the cubic $x^3+1=0$ and then the roots will be $-1$, $-w$, $-w^{2}$ where $w$ is the cube root of unity, so $a$ and $b$ will be equal to $-w$, and $-w^{2}$.
I wanted to solve this using a method which did not involve complex numbers but couldn't think of any, can anybody please help me.
On
Take root $a$ for example and note that $\,a^2-a+1=0\,$ since it's a root of the equation. Then:
$$a^2 = a-1 \tag{1}$$
$$a^2 - a = -1 \tag{2}$$
Multiply $(1)$ by $\,a \ne 0\,$ and use $(2)\,$ to get:
$$ a^3 = a^2-a = -1 $$
It follows that $a^{2009}=\left(a^3\right)^{669} \cdot a^2=(-1)^{669} \cdot a^2=-a^2\,$. Same goes for b, so the sum is:
$$-a^2-b^2=-a+1-b+1=-(a+b)+2=1$$
On
$$x^2 - x + 1 = 0 \implies x^3 - x^2 + x = 0$$
But $$x^2 - x = -1$$ (using the quadaratic again)
Thus $$x^3 = -1$$
So $a^3 = -1$ and $b^3 = -1$.
Thus $$a^{2009} + b^{2009} = a^{3 \cdot 669} a^2 + b^{3 \cdot 669} b^2 = -a^2 - b^2 = 2ab - (a + b)^2 $$
($2007 = 3\cdot669$)
But $a+b = 1$ (sum of roots of quadratic)
and $ab = 1$ (product of roots of quadratic)
Thus the answer is $$1$$
On
Let $k_n = a^n+b^n$ for integer $n$. Consider the proposition
$$k_n = a^n + b^n = 2\cos\frac{2\pi n}6$$
For $n = 0$, $a^0 + b^0 = 2 = 2\cos 0$.
For $n = 1$, $a + b = 1 = 2\cos \frac{2\pi}6$.
Assume the proposition is true for $m-1$ and $m$, where $m\ge 1$ is an integer,
$$k_i = a^i + b^i = 2\cos \frac{2\pi i}6,\quad i = m-1, m$$
For $n = m + 1$,
$$\begin{align*} a^{m+1} + b^{m+1} &= a(a^m+b^m) + b(a^m+b^m) - ab^m - a^mb\\ &= (a+b)(a^m+b^m) - ab(a^{m-1}+b^{m-1})\\ &= k_m - k_{m-1}\\ &= 2\cos\frac{2\pi m}6 - 2\cos\frac{2\pi(m-1)}6\\ &= 2\cdot(-2)\sin\left(\frac{2\pi}{6}\cdot\frac{2m-1}{2}\right)\sin\left(\frac{2\pi}{6}\cdot\frac{1}{2}\right)\\ &= -2\sin\left[\frac{2\pi}{6}\left(m -\frac{1}{2}\right)\right]\\ &= 2\cos\left[\frac {2\pi}6\left(m -\frac{1}{2}\right) +\frac{2\pi}{4}\right]\\ &= 2\cos\left[\frac {2\pi}6\left(m -\frac{1}{2}\right) +\frac{2\pi}{6}\cdot\frac{3}{2}\right]\\ &= 2\cos\frac {2\pi(m+1)}6 \end{align*}$$
So the proposition is true by induction for non-negative integers $n$, and
$$a^{2009}+b^{2009} = 2\cos\frac{2\pi\cdot2009}6 = 2\cos\frac{2\pi\cdot5}6 = 1$$
You can't choose the roots of a polynomial . There may be no roots in the real numbers. if you look at the equation you can get $$x^2-x=-1 \implies (x-1)x=-1 \implies \text {x is negative or x-1 is negative}$$ if x is negative so is x-1 so that's out, as a negative times a negative is a positive. x-1 being negative shows that x is in the half open interval from 0 to 1, x=0 implies 0=-1 , which is false. x=1 causes both x and x-1 to be non-negative (slightly different from positive, and x=1 wouldn't be in the half open interval). if x is rational then we have that the absolute values of x and x-1 are flipped fractions ( aka $x={a\over b}, x-1 = -{b\over a}$). otherwise the solutions would have to be irrational ( which then begs the question can two irrational numbers multiplied together equal a rational). if this last ditch effort fails ( admittedly I don't know) then there would be no solutions in the reals ( the set union of the rationals and irrationals)