Find the square root of 1595.69.

Step-by-step solution

1. 1

   Bracket the answer between two integers. Since $39^2=1521$ and $40^2=1600$, and $1595.69$ lies between them — much closer to $1600$ — the root is a little under $40$.

2. 2

   Refine using the difference from 40. Write the root as $40-d$. Then $(40-d)^2=1600-80d+d^2$, and we need this to equal $1595.69$, i.e. $80d-d^2=4.31$. Ignoring the tiny $d^2$ gives $d\approx\tfrac{4.31}{80}=0.0539$, so the root is about $40-0.0539=39.946$.

3. 3

   Bracket to two decimals with exact squares. Compute carefully, using $(40-t)^2=1600-80t+t^2$:

   $39.94^2=1600-4.8+0.0036=1595.2036,\qquad 39.95^2=1600-4+0.0025=1596.0025$

   Since $1595.2036<1595.69<1596.0025$, the root lies between $39.94$ and $39.95$.

4. 4

   Correct a bracketing that is often quoted wrongly. Values of $1595.0025$ for $39.95^2$ and $1595.8016$ for $39.96^2$ appear in some worked solutions — each is exactly $1$ too small. The true values are $1596.0025$ and $1596.8016$, which is why the correct bracket is $[39.94,39.95]$ and not $[39.95,39.96]$.

5. 5

   State the value to four decimal places.

   $\sqrt{1595.69}=39.9460887\ldots\approx39.9461$

   This sits inside the corrected bracket ✓ and matches the linear estimate of $39.946$ from step 2 ✓.

6. 6

   Verify by squaring back. $39.9461^2=1595.6909$, which overshoots $1595.69$ by only $0.0009$; $39.946^2=1595.6829$, undershooting by $0.0071$. So $39.9461$ is the better four-decimal rounding ✓. The value is irrational, since $\sqrt{1595.69}=\tfrac{\sqrt{159569}}{10}$ and $159569$ is prime, hence certainly not a perfect square, so no exact decimal exists.