SOLUTION: tough one! One fourth of a herd of camels was seen in the forest, twice the square root of that herd had gone to the mountain slope and 3 times 5 camels remained on the river ba

Algebra ->  Radicals -> SOLUTION: tough one! One fourth of a herd of camels was seen in the forest, twice the square root of that herd had gone to the mountain slope and 3 times 5 camels remained on the river ba      Log On


   



Question 81547: tough one!
One fourth of a herd of camels was seen in the forest, twice the square root of that herd had gone to the mountain slope and 3 times 5 camels remained on the river bank. WHat is the numerical measure of that herd of camels?
so far I have:
h= herd
h/4 = 2 sq root h + (3.5) = h

Found 2 solutions by doctor_who, bucky:
Answer by doctor_who(15) About Me  (Show Source):
You can put this solution on YOUR website!
Are you sure you copied that down right ? Three times five camels is fifteen camels, not three and a half.
Do a quick double-check and let me know if that is still right, thanks.

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Your work:
.
h= herd <--- OK
h/4 = 2 sq root h + (3.5) = h <--- a couple of comments here. I think you meant the equal
sign between h/4 and 2sq root h to be a + sign. At least it should be +. And I think
you meant the 3.5 to be 3 times 5. Making those changes results in:
.
h%2F4+%2B+2%2Asqrt%28h%29+%2B+%283%2A5%29+=+h
.
Multiply out the 3*5 to 15 and put that into the equation:
.
h%2F4+%2B+2%2Asqrt%28h%29+%2B+15+=+h
.
Now what you want to do is to get the term containing sqrt%28h%29 by itself on one side
of the equation and everything else on the other side. Start by getting rid of the 15
on the left side by subtracting 15 from both sides to get:
.
h%2F4+%2B+2%2Asqrt%28h%29+=+h+-+15
.
Then get rid of the h%2F4 on the left side by subtracting h%2F4 from both sides
to get:
.
2%2Asqrt%28h%29=+h+-+h%2F4+-+15
.
The terms on the right side that contain h can be combined by subtraction to get:
.
2%2Asqrt%28h%29+=+3h%2F4+-+15
.
and by multiplying the 15 on the right side by 4%2F4 you get:
.
2%2Asqrt%28h%29+=+%283h+-+60%29%2F4
.
Now get rid of the denominator on the right side by multiplying both sides of the equation
by 4 to get:
.
8%2Asqrt%28h%29+=+3h+-+60
.
Now square both sides of this equation. When you do it becomes:
.
64h+=+9h%5E2+-+360h+%2B+3600
.
Subtract 64h from both sides to get the standard quadratic form:
.
0+=+9h%5E2+-+424h+%2B+3600
.
or in the more conventional form (transposed):
.
9h%5E2+-+424h+%2B+3600+=+0
.
The left side of this equation factors to make the equation become:
.
%289h+-+100%29%2A%28h+-+36%29=+0
.
This equation will be true if either of the factors on the left side is zero. By
setting the
first factor equal to zero you get:
.
9h+-+100+=+0
.
Add 100 to both sides and it becomes:
.
9h+=+100
.
And dividing both sides by 9 you find:
.
h+=+100%2F9
.
Since 9 does not divide into 100 evenly, discard this answer. The herd should have a whole
number of members.
.
Set the second factor equal to zero ...
.
h+-+36+=+0
.
add 36 to both sides and you get:
.
h+=+36
.
Let's try it:
.
a quarter of the herd is 9. The square root of the herd is 6 and twice that is 12. And
finally 15 members of the herd ... the total is 9 + 12 + 15 = 36. Yup ... that works.
.
The herd consists of 36 animals.
.
Note ... the factoring above took a little effort. You could have used the quadratic
formula just as easily to find that h = 36 and h = 100/9 were the two possible answers.
.
Hope this helps you with this problem and with understanding how you can get a solution.