Question

What two integers does sqrt41 fall between? explain reasoning

Answer 1

Answer:

Below!

Step-by-step explanation:

Given square root:

$\sqrt41$

Let's note a few perfect square roots.

$1 = \sqrt{1 \times 1} = \sqrt{1}$

$2 = \sqrt{2 \times 2} = \sqrt{4}$

$3 = \sqrt{3 \times 3} =\sqrt{9}$

$4 = \sqrt{4 \times 4} =\sqrt{16}$

$5 = \sqrt{5 \times 5} =\sqrt{25}$

$6 = \sqrt{6 \times 6} =\sqrt{36}$

$7 = \sqrt{7 \times 7} =\sqrt{49}$

$8 = \sqrt{8 \times 8} =\sqrt{64}$

And so on...

When looking at the perfect square roots we identified, we can say that:

$\sqrt{36} < \sqrt{41} < \sqrt{49}$

Therefore,

$6 < \sqrt{41} < 7$

We can conclude that $\sqrt{41}$ is between 6 and 7.

Answer 2

Let's check the squares of first 10 number

• 1²=1
• 2²=4
• 3²=9
• 4²=16
• 5²=25
• 6²=36
• 7²=49
• 8²=64
• 9²=81
• 10²=100

41 lies between 36 and 49

Hence

• √41 lies in between 6 and 7