The problem asks to find two consecutive integers that bound the square root of 17, $\sqrt{17}$. In other words, we need to find integers $a$ and $a+1$ such that $a < \sqrt{17} < a+1$.

1. Problem Description

The problem asks to find two consecutive integers that bound the square root of 17, $\sqrt{17}$. In other words, we need to find integers $a$ and $a+1$ such that $a < \sqrt{17} < a+1$.

2. Solution Steps

We need to find the perfect squares that are close to

1

7. $1^2 = 1$

$2^2 = 4$

$3^2 = 9$

$4^2 = 16$

$5^2 = 25$

We see that 16 is less than 17 and 25 is greater than

1

7. So, we have $16 < 17 < 25$.

Taking the square root of each term, we get $\sqrt{16} < \sqrt{17} < \sqrt{25}$.

Since $\sqrt{16} = 4$ and $\sqrt{25} = 5$, we have $4 < \sqrt{17} < 5$.

The two consecutive integers are 4 and

5.

3. Final Answer

4 < $\sqrt{17}$ < 5