√(4K)^2 = √1849, then K=
Answers
Answer:
We could first seek to find the prime factorisation of
1849
, but as we shall see it is actually the square of a prime number, so that would be somewhat tedious.
Alternatively, let's split it into pairs of digits from the right to get:
18
|
49
Examining the leading
18
, note that it lies between
4
2
and
5
2
:
4
2
=
16
<
18
<
25
=
5
2
So:
4
<
√
18
<
5
and hence:
40
<
√
1849
<
50
To find a suitable correction, we can linearly interpolate between
40
and
50
to find:
√
1849
≈
40
+
(
50
−
40
)
⋅
1849
−
40
2
50
2
−
40
2
√
1849
≈
40
+
10
⋅
1849
−
1600
2500
−
1600
√
1849
≈
40
+
2490
900
√
1849
≈
40
+
2.49
+
0.249
+
0.0249
+
...
√
1849
≈
42.76
Hmmm... That's close to
43
, let's try
43
2
...
43
⋅
43
=
40
2
+
2
⋅
40
⋅
3
+
3
2
=
1600
+
240
+
9
=
1849
So:
√
1849
=
43
Answer:
sol is 3 since √(4k) ^2=√1849
Step-by-step explanation:
the square and square root cancled then
4k=√(43)^2
again saquare square root cancled
then4k=43
then k=3