Radical Equations

[CodePoet]

Hi, I'm having issues with a word problem that I'm assuming is based off of radical equations. The title of the section was "other equations"; I believe this particular problem has to do with radicals as the pre-problem instruction was to use the Pythagorean formula.
Here is the text of the word problem.
"Two vertical poles of lengths 6 feet and 8 feet respectively stand 10 feet apart. A cable reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole. Where should this point be located to use 18 feet of cable?"
The given picture shows two right triangles that are touching at one point with their bases forming a continuous line, somewhat like this, assuming the characters would meet to form a full triangle on either side
6|_\/_|8
The length of the base of the triangle with the 6 foot pole (the first underscore in the rather horrible but only depiction I could type figure) is x. The length of the base of the triangle with the 8 foot pole (second underscore) is 10-x.
Based on the description and the picture in the textbook, the 18 feet of cable forms the hypotenuses of the two triangles. I know the hypotenuse of a triangle (c) can be found with the Pythagorean theorem
a[sup:2okgs0ij]2[/sup:2okgs0ij]+b[sup:2okgs0ij]2[/sup:2okgs0ij]=c[sup:2okgs0ij]2[/sup:2okgs0ij] which I then solved for c to be
c=sqrt(a[sup:2okgs0ij]2[/sup:2okgs0ij]+b[sup:2okgs0ij]2[/sup:2okgs0ij]). Based on that formula and the knowledge that the length of both hypotenuses was equal to 18 feet, I came up with the equation
18=sqrt(x[sup:2okgs0ij]2[/sup:2okgs0ij]+36)+sqrt[(10-x)[sup:2okgs0ij]2[/sup:2okgs0ij]+64]
However, when I attempt to solve this problem I end up with very impossible answers. I'd like to perhaps know if my equation is wrong or if it isn't, the first couple of steps to solve as perhaps my process was wrong. I cannot post my work on this problem as I ended up tearing the sheet of paper up in frustration. Thank you for any help you can give.

 

[Deleted member 4993]

Hi, I'm having issues with a word problem that I'm assuming is based off of radical equations. The title of the section was "other equations"; I believe this particular problem has to do with radicals as the pre-problem instruction was to use the Pythagorean formula.
Here is the text of the word problem.
"Two vertical poles of lengths 6 feet and 8 feet respectively stand 10 feet apart. A cable reaches from the top of one pole to some point on the ground between the poles and then to the top of the other pole. Where should this point be located to use 18 feet of cable?"
The given picture shows two right triangles that are touching at one point with their bases forming a continuous line, somewhat like this, assuming the characters would meet to form a full triangle on either side
6|_\/_|8
The length of the base of the triangle with the 6 foot pole (the first underscore in the rather horrible but only depiction I could type figure) is x. The length of the base of the triangle with the 8 foot pole (second underscore) is 10-x.
Based on the description and the picture in the textbook, the 18 feet of cable forms the hypotenuses of the two triangles. I know the hypotenuse of a triangle (c) can be found with the Pythagorean theorem
a[sup:ccxsvuyw]2[/sup:ccxsvuyw]+b[sup:ccxsvuyw]2[/sup:ccxsvuyw]=c[sup:ccxsvuyw]2[/sup:ccxsvuyw] which I then solved for c to be
c=sqrt(a[sup:ccxsvuyw]2[/sup:ccxsvuyw]+b[sup:ccxsvuyw]2[/sup:ccxsvuyw]). Based on that formula and the knowledge that the length of both hypotenuses was equal to 18 feet, I came up with the equation
18=sqrt(x[sup:ccxsvuyw]2[/sup:ccxsvuyw]+36)+sqrt[(10-x)[sup:ccxsvuyw]2[/sup:ccxsvuyw]+64]
However, when I attempt to solve this problem I end up with very impossible answers. I'd like to perhaps know if my equation is wrong or if it isn't, the first couple of steps to solve as perhaps my process was wrong. I cannot post my work on this problem as I ended up tearing the sheet of paper up in frustration. Thank you for any help you can give.

your equation looks correct to me.

sqrt[(10-x)[sup:ccxsvuyw]2[/sup:ccxsvuyw]+64] = 18 - sqrt(x[sup:ccxsvuyw]2[/sup:ccxsvuyw]+36)

164 - 20x + x[sup:ccxsvuyw]2[/sup:ccxsvuyw] = 18[sup:ccxsvuyw]2[/sup:ccxsvuyw] + x[sup:ccxsvuyw]2[/sup:ccxsvuyw] + 36 - 36sqrt(x[sup:ccxsvuyw]2[/sup:ccxsvuyw]+36)

-20x +164-360 = - 36sqrt(x[sup:ccxsvuyw]2[/sup:ccxsvuyw]+36)

36sqrt(x[sup:ccxsvuyw]2[/sup:ccxsvuyw]+36) = 196 + 20 x

Now square again - simplify - solve quadratic - check the solution...

 

[CodePoet]

your equation looks correct to me.

sqrt[(10-x)[sup:37bl3hq2]2[/sup:37bl3hq2]+64] = 18 - sqrt(x[sup:37bl3hq2]2[/sup:37bl3hq2]+36)

164 - 20x + x[sup:37bl3hq2]2[/sup:37bl3hq2] = 18[sup:37bl3hq2]2[/sup:37bl3hq2] + x[sup:37bl3hq2]2[/sup:37bl3hq2] + 36 - 36sqrt(x[sup:37bl3hq2]2[/sup:37bl3hq2]+36)

-20x +164-360 = - 36sqrt(x[sup:37bl3hq2]2[/sup:37bl3hq2]+36)

36sqrt(x[sup:37bl3hq2]2[/sup:37bl3hq2]+36) = 196 + 20 x

Now square again - simplify - solve quadratic - check the solution...

I'm not sure where I'm going wrong with my work....here I do not divide out the 36 right away because I tried doing that in my last iteration and failed to come up with even a reasonably wrong answer. Here is the current method that's on my paper.
1296x[sup:37bl3hq2]2[/sup:37bl3hq2]+36=38416+7840x+400x[sup:37bl3hq2]2[/sup:37bl3hq2]
896x[sup:37bl3hq2]2[/sup:37bl3hq2]-7840x-38380=0
4(224x[sup:37bl3hq2]2[/sup:37bl3hq2]-1960x-9595)=0
From here with the large numbers I decided to use the quadratic formula
1960(plus or minus)sqrt[3841600-4(224)(-9595)]/448
which ended up as
12.25 which is impossible, as going 12.25 feet past the base of the 6 foot pole (see first post) would be past the 8 foot pole. The answer given by the text is "either 1.2 feet or 7.5 feet from the base of the 6 foot pole" which I have not gotten remotely close to no matter how I work the problem. Any further guidance would be appreciated.

 

[mmm4444bot]

… 36 sqrt(x[sup:30ozmna3]2[/sup:30ozmna3] + 36) = 196 + 20x

Now square again …

I'm not sure where I'm going wrong with my work …

1296x[sup:30ozmna3]2[/sup:30ozmna3] + 36 = 38416 + 7840x + 400x[sup:30ozmna3]2[/sup:30ozmna3] 36 is not correct.

Check your use of the Distributive Property, on the lefthand side, after squaring.

 

[CodePoet]

STILL coming up with an impossible answer...hope I'm not being troublesome at this point....
1296x[sup:sqwn22py]2[/sup:sqwn22py]+1296=38416+7840x+400x[sup:sqwn22py]2[/sup:sqwn22py]
896x[sup:sqwn22py]2[/sup:sqwn22py]-7840x-37120=0
32(28x[sup:sqwn22py]2[/sup:sqwn22py]-245x-1160)=0
Then attempting to solve with the quadratic equation again....
{245(plus or minus)sqrt[60025-4(28)(-1160)]}/56
Which gave me around 12.16 and -3.41 neither of which are possible, much less correct.

 

[Mrspi]

STILL coming up with an impossible answer...hope I'm not being troublesome at this point....
1296x[sup:37rc3e27]2[/sup:37rc3e27]+1296=38416+7840x+400x[sup:37rc3e27]2[/sup:37rc3e27]
896x[sup:37rc3e27]2[/sup:37rc3e27]-7840x-37120=0
32(28x[sup:37rc3e27]2[/sup:37rc3e27]-245x-1160)=0
Then attempting to solve with the quadratic equation again....
{245(plus or minus)sqrt[60025-4(28)(-1160)]}/56
Which gave me around 12.16 and -3.41 neither of which are possible, much less correct.

Ok....let's start with the equation you were told was correct:

36 * sqrt(x[sup:37rc3e27]2[/sup:37rc3e27] + 36) = 196 + 20x

Square both sides to eliminate the radical:

[36 * sqrt(x[sup:37rc3e27]2[/sup:37rc3e27] + 36)][sup:37rc3e27]2[/sup:37rc3e27] = (196 + 20x)[sup:37rc3e27]2[/sup:37rc3e27]

36[sup:37rc3e27]2[/sup:37rc3e27] * [sqrt(x[sup:37rc3e27]2[/sup:37rc3e27] + 36)][sup:37rc3e27]2[/sup:37rc3e27] = (196 + 20x)[sup:37rc3e27]2[/sup:37rc3e27]

1296 * (x[sup:37rc3e27]2[/sup:37rc3e27] + 36) = 38416 + 7840x + 400x[sup:37rc3e27]2[/sup:37rc3e27]

1296x[sup:37rc3e27]2[/sup:37rc3e27] + 46656 = 38416 + 7840x + 400x[sup:37rc3e27]2[/sup:37rc3e27]

896x[sup:37rc3e27]2[/sup:37rc3e27] - 7840x + 8240 = 0

We can divide both sides by 16....

56x[sup:37rc3e27]2[/sup:37rc3e27] - 490x + 515 = 0

NOW try the quadratic formula....

 

[Denis]

Perhaps a less confusing way is assign x to other leg and y to hypotenuse of right triangle with side = 6.
Then the other leg of the side 8 triangle = 10-x and hypotenuse = 18-y.

side 6 triangle:
y^2 = x^2 + 36
y^2 - x^2 = 36 [1]

side 8 triangle:
(18 - y)^2 = (10 - x)^2 + 64 ; simplify to get:
y^2 - x^2 = 36y - 20x - 160 [2]

So, from [1],[2]:
36y - 20x - 160 = 36 ; simplify:
y = (5x + 49) / 9 [3]

Substitute [3] in [1] (the easiest!) which will lead quickly to 56x^2 - 490x + 515 = 0

Solve to get 2 solutions (rounded): x = 7.53, y = 9.63 and x = 1.22, y = 6.12

It's always a good idea to make life "easier!" by 1st trying a case that YOU make up,
using 2 known right triangles; as example: 3-4-5 and 5-12-13.

The initial problem would then be:
Two vertical poles of lengths 3 feet and 5 feet respectively stand 16 feet apart.
A cable reaches from the top of one pole to some point on the ground between the poles and then
to the top of the other pole. Where should this point be located to use 18 feet of cable?"

Now you try the problem "knowing" what you have to end up with; get my drift?
By the way, you'll get an extra solution again.

 

[CodePoet]

Thank you everyone, I ended up using mrspi's method since that was closest to what I started with, but I'll be sure to keep your insights in mind for the future Denis.

 

[mmm4444bot]

STILL coming up with an impossible answer …

1296x[sup:2xg4xt7x]2[/sup:2xg4xt7x] + 1296 = 38416 + 7840x + 400x[sup:2xg4xt7x]2[/sup:2xg4xt7x] Still not properly distributing 1296 across the sum x^2 + 36.

As Mrs. Pi shows, 1296*36 = 46656, not 1296 .

I'm hoping, Code Poet, that you recognized this from studying Mrs. Pi's post (as opposed to mindless duplication). Recognizing when and how to apply the Distributive Property is paramount.