Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \ne 1.$  Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6.$  Then $\alpha$ and $\beta$ satisfy the quadratic
\[x^2 + ax + b = 0\]for some real numbers $a$ and $b.$  Enter the ordered pair $(a,b).$
From the equation $\omega^7 = 1,$ $\omega^7 - 1 = 0,$ which factors as
\[(\omega - 1)(\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) = 0.\]Since $\omega \neq 1,$
\[\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1 = 0.\]We have that
\[\alpha + \beta = \omega + \omega^2 + \omega^4 + \omega^3 + \omega^5 + \omega^6 = -1.\]Also,
\begin{align*}
\alpha \beta &= (\omega + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6) \\
&= \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10} \\
&= \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega + 1 + \omega^2 + \omega^3 \\
&= 2 + (\omega^6 + \omega^5 + \omega^4 + \omega^3 + \omega^2 + \omega + 1) \\
&= 2.
\end{align*}Then by Vieta's formulas, $\alpha$ and $\beta$ are the roots of $x^2 + x + 2 = 0,$ so $(a,b) = \boxed{(1,2)}.$
(1,2)
Level 5
Intermediate Algebra
Problem: Let $\omega$ be a complex number such that $\omega^7 = 1$ and $\omega \ne 1.$  Let $\alpha = \omega + \omega^2 + \omega^4$ and $\beta = \omega^3 + \omega^5 + \omega^6.$  Then $\alpha$ and $\beta$ satisfy the quadratic
\[x^2 + ax + b = 0\]for some real numbers $a$ and $b.$  Enter the ordered pair $(a,b).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the values of \( a \) and \( b \) in the quadratic equation satisfied by \( \alpha \) and \( \beta \), we first need to compute \( \alpha \) and \( \beta \) explicitly. 

Since \( \omega \) is a primitive 7th root of unity, we have the roots \( \omega, \omega^2, \omega^3, \omega^4, \omega^5, \omega^6 \) corresponding to the equation \( x^7 - 1 = 0 \). The sum of all roots is zero, thus:

\[
1 + \omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = 0.
\]

This implies:

\[
\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6 = -1.
\]

Now, we can express \( \alpha \) and \( \beta \):

\[
\alpha = \omega + \omega^2 + \omega^4,
\]
\[
\beta = \omega^3 + \omega^5 + \omega^6.
\]

Next, we can find \( \alpha + \beta \):

\[
\alpha + \beta = (\omega + \omega^2 + \omega^4) + (\omega^3 + \omega^5 + \omega^6) = -1.
\]

Now, we compute \( \alpha \beta \):

\[
\alpha \beta = (\omega + \omega^2 + \omega^4)(\omega^3 + \omega^5 + \omega^6).
\]

Expanding this product, we can group terms based on the powers of \( \omega \):

\[
= \omega \cdot \omega^3 + \omega \cdot \omega^5 + \omega \cdot \omega^6 + \omega^2 \cdot \omega^3 + \omega^2 \cdot \omega^5 + \omega^2 \cdot \omega^6 + \omega^4 \cdot \omega^3 + \omega^4 \cdot \omega^5 + \omega^4 \cdot \omega^6.
\]

Calculating each term, we have:

\[
= \omega^4 + \omega^6 + \omega^7 + \omega^5 + \omega^7 + \omega^8 + \omega^7 + \omega^9 + \omega^{10}.
\]

Since \( \omega^7 = 1 \), we can reduce the powers:

\[
= \omega^4 + \omega^6 + 1 + \omega^5 + 1 + \omega + 1 + \omega^2 + \omega^3.
\]

Combining like terms, we find:

\[
= 3 + (\omega + \omega^2 + \omega^3 + \omega^4 + \omega^5 + \omega^6) = 3 - 1 = 2.
\]

Thus, we have:

\[
\alpha + \beta = -1,
\]
\[
\alpha \beta = 2.
\]

The quadratic equation with roots \( \alpha \) and \( \beta \) can be expressed as:

\[
x^2 - (\alpha + \beta)x + \alpha \beta = 0.
\]

Substituting the values we found:

\[
x^2 + x + 2 = 0.
\]

From this, we identify \( a = 1 \) and \( b = 2 \). Therefore, the ordered pair is:

(1, 2)<|END_OF_LATENT|> The final answer is $(1,2)$.