### Rationals Between...

What fractions can you find between the square roots of 65 and 67?

### Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

### Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

# Roots Near 9

##### Age 14 to 16 Short Challenge Level:

Considering values of $n$
If $n=81$, then $\sqrt n=9$, so numbers close to $81$ will work.

What other square numbers are close to $81$? We know that $\sqrt{64}=8$, so $n$ can't be $64$. But if $n$ is a little bit larger than $64$, then $\sqrt n$ will be just larger than $8$, so the distance between $\sqrt n$ and $9$ will be less than $1$.

We also know that $\sqrt{100}=10$, so $n$ can't be $100$. But if $n$ is a little bit smaller than $100$, then $\sqrt n$ will be just smaller than $10$, so the distance between $\sqrt n$ and $9$ will be less than $1$.

So $n$ can be any number from $65$ to $99$, which is a total of $35$ numbers.

Creating inequalities
If $\sqrt n < 9$, then we must have $9-\sqrt n<1$.
If $\sqrt n > 9$, then we must have $\sqrt n-9<1$.
Solving these two inequalities will give us the possible values of $n$.
\begin{align} 9-\sqrt n &<1\\ \Rightarrow 9&<1+\sqrt n\\ \Rightarrow 8&<\sqrt n\end{align} Because $8$ and $\sqrt n$ are both positive, we can square both sides without risking multiplying by a negative number (which would change the inequality sign!). So $$8^2<\sqrt n^2\Rightarrow 64<n$$

Similarly, we can solve the other inequality:$$\sqrt n -9 <1\Rightarrow \sqrt n<10$$ Again, since $10$ and $\sqrt n$ are both positive, we can square both sides without risking multiplying by a negative number: $$\sqrt n^2<10^2\Rightarrow n<100$$

So $n$ is greater than $64$ but less than $100$, which means that $n$ can be any number from $65$ to $99$. That is a total of $35$ numbers.
You can find more short problems, arranged by curriculum topic, in our short problems collection.