Irrational Roots
You're used to working with quadratics with integer roots, but what about when the roots are irrational?
Age
16 to 18
Challenge level
Problem
The roots of quadratic polynomials can be nice, integer values. For example $x^2+4x+3$ has $x=-3$ as a root. However, this is not always the case. You will have encountered many quadratic polynomials with roots that are fractions or even irrational numbers.
Convince yourself that $x=\sqrt{2}$ is a root of the quadratic equation $x^2-2=0$ and that $x=\sqrt{3}$ is a root of the quadratic equation $x^2-3=0$.
- Can you find a quadratic polynomial with integer coefficients which has $x=1+\sqrt{2}$ as a root?
What is the other root of this polynomial?
- What if instead $x=1+\sqrt{3}$ is a root? What would the quadratic polynomial be now?
What would the other root be this time?
- Can you generalise your answers to the case where $1+\sqrt{n}$ is a root?
What about the case where $m+\sqrt{n}$ is a root?
This is an Underground Mathematics resource.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Underground Mathematics is hosted by Cambridge Mathematics. The project was originally funded by a grant from the UK Department for Education to provide free web-based resources that support the teaching and learning of post-16 mathematics.
Visit the site at undergroundmathematics.org to find more resources, which also offer suggestions, solutions and teacher notes to help with their use in the classroom.
Getting Started
I know that the quadratic factorises as $(x-(1+\sqrt{2}))$ times
something. So perhaps I could write the quadratic as
$$(x-(1+\sqrt{2}))(ax+b)$$
and work out what $a$ and $b$ could be.
Alternatively, perhaps I could think about the relation between
$x=\sqrt{2}$ and $x^2-2=0$, and see where that leads me to if I start
with $x=1+\sqrt{2}$ instead.
As a third possible approach, I could think about the graph of
$y=x^2-2$, which has $x=\sqrt{2}$ as a solution, and wonder what I
would have to do to get $x=1+\sqrt{2}$ as a solution instead.
something. So perhaps I could write the quadratic as
$$(x-(1+\sqrt{2}))(ax+b)$$
and work out what $a$ and $b$ could be.
Alternatively, perhaps I could think about the relation between
$x=\sqrt{2}$ and $x^2-2=0$, and see where that leads me to if I start
with $x=1+\sqrt{2}$ instead.
As a third possible approach, I could think about the graph of
$y=x^2-2$, which has $x=\sqrt{2}$ as a solution, and wonder what I
would have to do to get $x=1+\sqrt{2}$ as a solution instead.
Student Solutions
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.