Package of problems

Solution

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?

Approach 1

We can approach this problem by thinking about the factorised form of the quadratic polynomial. If $x=1+\sqrt{2}$ is a root then we can factorise the original quadratic as $(x-(1+\sqrt{2}))(ax+b)$.

If we expand the factorised form we get $ax^2+(b-a(1+\sqrt{2}))x-b(1+\sqrt{2}).$

Now, we know that we are looking for a quadratic with integer coefficients so we know that the constant term, $-b(1+\sqrt{2})$ must be an integer, say $n$. If $-b(1+\sqrt{2})=n$, then $b=-\frac{n}{1+\sqrt{2}}.$ We can rationalise the denominator by multiplying by $1-\sqrt{2}$ to get

\begin{align*} b &{}=-\frac{n(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})} \\ &{}=-\frac{n(1-\sqrt{2})}{1-2} \\ &{}=n(1-\sqrt{2}). \end{align*}

We are making use of the difference of two squares here. $(a+b)(a-b)=a^2-b^2$

This gives us a constant term of $n$.

We also require that the coefficient of $x$, $b-a(1+\sqrt{2})$ is an integer. Substituting our value for $b$ in here gives $n(1-\sqrt{2})-a(1+\sqrt{2})$.

For this to be an integer we require that $a=-n$ so that the surds cancel $n(1-\sqrt{2})+ n(1+\sqrt{2})=2n.$

So we have a quadratic polynomial of the form $-nx^2+2nx+n$ or equivalently, $nx^2-2nx-n$. In the simplest case this is $x^2-2x-1$.

The other root of the polynomial will be $x=1-\sqrt{2}$.

Approach 2

When we think about the first example, $x^2-2=0$, we see that it is just $x=\sqrt{2}$ squared and then rearranged.

So if we start with $x=1+\sqrt{2}$, we could just square it, but then we get $x^2=3+2\sqrt{2}$, which hasn’t really helped us.

But if we rearrange it first to get the square root alone on one side, we have $x-1=\sqrt{2}.$ This can now be squared to get $(x-1)^2=2,$ which does the job: expanding the brackets gives $x^2-2x+1=2$, so $x^2-2x-1=0$ has $x=1+\sqrt{2}$ as a root.

Working backwards, we see that $(x-1)^2=2$ gives $x-1=\pm\sqrt{2}$, so the other root is $x=1-\sqrt{2}$.

Approach 3

When we think about the graph of $y=x^2-2$, a parabola, we see that it has a root at $x=\sqrt{2}$ and another one at $x=-\sqrt{2}$. As we want to have a root at $x=1+\sqrt{2}$, we can translate this parabola $1$ unit to the right so that it now has a root at $\sqrt{2}+1$. This can be achieved if we remember that $f(x)\rightarrow f(x-a)$ corresponds to a horizontal translation $a$ to the right. In this case, we begin with $f(x)=x^2-2$ so that $f(x-1)=(x-1)^2-2=x^2-2x-1$.

Therefore a quadratic polynomial with $x=1+\sqrt{2}$ as a root is $x^2-2x-1$, as with the other two approaches.

The second root of $f(x-1)$ is $-\sqrt{2}+1$, as the second root $-\sqrt{2}$ of the original polynomial is also translated $1$ unit to the right.

The second root could also have been found using the quadratic formula.

Approach 4

We have $x=1+\sqrt{2}$. When we square this, we get $x^2=3+2\sqrt{2}$. This might remind us of one method of solving simultaneous equations, where we eliminate variables one at a time.

In this case, if we subtract $2x$ from $x^2$, we’ll eliminate the $\sqrt{2}$: $x^2-2x=(3+2\sqrt{x})-2(1+\sqrt{x})=1,$ so $x^2-2x-1=0$, and we have our required polynomial.

We can now find the other root of this quadratic by any standard technique, though it is not clear from this approach what the other root will be.

Which of these four approaches do you prefer?

Which of these four approaches can be most easily adapted to other situations?

What if instead $x=1+\sqrt{3}$ is a root? How would the quadratic polynomial look now? What would the other root be this time?

This is a very similar question to that above. It makes sense that we can use the same approaches.

They all give the answer $x^2-2x-2=0$, and the other root is $1-\sqrt{3}$.

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?

Think about how the solutions to the first two questions could inform your answer here. Perhaps think about some other particular cases if you are unsure.