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Aneesh from West Island School has given a very nice solution to the general problem of finding the other root of a quadratic polynomial with integer coefficients which has $m+\sqrt{n}$ as one of the roots:
I have chosen to write an answer attempting to provide the solution for the
general case, when one of the roots is "$m+√n$".
Consider the general order two polynomial with integer coefficients a, b
and c.
This is $ax^2 + bx + c$. We know that this order two polynomial must have two
complex roots. One root "$m+\sqrt{n}$" has been given, and thus the
other root may not have an imaginary part. (Otherwise upon expansion, our
order two polynomial will most certainly not have integer coefficients).
Thus we may re-write our order two polynomials entirely in terms of its
roots. This is shown below:
$$ax^2 + bx + c = a(x-r1)(x-r2)$$
$a$ may not be equal to zero, or else our
polynomial will not be order two, so we may divide both sides of our
equality by $a$. Producing:
$$(1) x^2 + (b/a)x + (c/a) = (x-r1)(x-r2)$$
We may now deduce that
$$(b/a) =-(r1+r2)$$
$(b/a)$ is a member of the rationals, so we must ensure $r1+r2$ is
a member of the rationals. We know $r1 = m+√n$, and so $r1+r2 = m+√n+r2$.
If $r2$ is in the rationals, $r1+r2$ is irrational. Contradiction. $r2$ must
belong to the irrationals. However, we may restrict the value of $r2$ even
further. $r2$ may ONLY be an expression in the form $k - \sqrt{n}$ where $k$ is in
the rationals. This is because we wish to discard the $\sqrt{n}$ term in our
$r1+r2$ expression so as to obtain a rational number. Good, but we still do
not know what our value of $k$ should be.
On looking at (1), it is easy to see that
$$r1r2 = c/a$$
The product of our roots must produce a rational number. We also know that $r1 = m+\sqrt{n}$ and $r2= k-\sqrt{n}$. On multiplying $r1$ by $r2$, we get,$ mk + \sqrt{n}(k-m) - n$. $m,k,n$ are
ALL rationals. Also we know that $r1r2$ must be rational. Thus we must not
have a $\sqrt{n}$ term in our expression (we assume n is non-square, otherwise
$(m+\sqrt{n})$ is simply an integer in which case the problem becomes trivial.)
and so $k-m = 0$. Thus $k=m$. So we now know that in fact, our second root must
be $r2 = m-\sqrt{n}$ . This gives us the quadratic
$$a(x-(m+\sqrt{n}))(x-(m-\sqrt{n}))$$
Thus $r2$ is simply the conjugate of $r1$ which is as expected.
Irrational Roots
Age 16 to 18
Challenge Level 
Aneesh from West Island School has given a very nice solution to the general problem of finding the other root of a quadratic polynomial with integer coefficients which has $m+\sqrt{n}$ as one of the roots:
I have chosen to write an answer attempting to provide the solution for the
general case, when one of the roots is "$m+√n$".
Consider the general order two polynomial with integer coefficients a, b
and c.
This is $ax^2 + bx + c$. We know that this order two polynomial must have two
complex roots. One root "$m+\sqrt{n}$" has been given, and thus the
other root may not have an imaginary part. (Otherwise upon expansion, our
order two polynomial will most certainly not have integer coefficients).
Thus we may re-write our order two polynomials entirely in terms of its
roots. This is shown below:
$$ax^2 + bx + c = a(x-r1)(x-r2)$$
$a$ may not be equal to zero, or else our
polynomial will not be order two, so we may divide both sides of our
equality by $a$. Producing:
$$(1) x^2 + (b/a)x + (c/a) = (x-r1)(x-r2)$$
We may now deduce that
$$(b/a) =-(r1+r2)$$
$(b/a)$ is a member of the rationals, so we must ensure $r1+r2$ is
a member of the rationals. We know $r1 = m+√n$, and so $r1+r2 = m+√n+r2$.
If $r2$ is in the rationals, $r1+r2$ is irrational. Contradiction. $r2$ must
belong to the irrationals. However, we may restrict the value of $r2$ even
further. $r2$ may ONLY be an expression in the form $k - \sqrt{n}$ where $k$ is in
the rationals. This is because we wish to discard the $\sqrt{n}$ term in our
$r1+r2$ expression so as to obtain a rational number. Good, but we still do
not know what our value of $k$ should be.
On looking at (1), it is easy to see that
$$r1r2 = c/a$$
The product of our roots must produce a rational number. We also know that $r1 = m+\sqrt{n}$ and $r2= k-\sqrt{n}$. On multiplying $r1$ by $r2$, we get,$ mk + \sqrt{n}(k-m) - n$. $m,k,n$ are
ALL rationals. Also we know that $r1r2$ must be rational. Thus we must not
have a $\sqrt{n}$ term in our expression (we assume n is non-square, otherwise
$(m+\sqrt{n})$ is simply an integer in which case the problem becomes trivial.)
and so $k-m = 0$. Thus $k=m$. So we now know that in fact, our second root must
be $r2 = m-\sqrt{n}$ . This gives us the quadratic
$$a(x-(m+\sqrt{n}))(x-(m-\sqrt{n}))$$
Thus $r2$ is simply the conjugate of $r1$ which is as expected.
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