Solving a Complex Math Problem: Real and Imaginary Solutions

In summary, the problem involves finding the height of a ladder leaning against a building and touching a 1m x 1m x 1m packing case at the bottom. The ladder is 10m long and the goal is to find how far up the wall it reaches. Using similar triangle theory and Pascal's formula, an initial solution of x = -1.88 and y = -0.53 is found, but this is incorrect. Using the fact that the ladder must touch the box, the equation (1/x) + (1/y) = 1 is derived, leading to the correct solution.
  • #1
Noir
27
0

Homework Statement


Apparantley the following problem can't be solved with a graphics calculator and we have to state why this is, and give all the possible solutions.

A ladder AB leans against the side of a building. Where the wall meets the ground is a cubic packing case 1m x 1m x 1m. The ladder is adjusted so that it rests on the ground, touches the packing case and rests against the wall. If the ladder is 10 metres long, how far up the wall will the ladder reach.

That was the question and a diagram was drawn, except I can't scan it in at the moment. Thinking on the problem there is more than one answer to the problem. Apparantley there are real and imaginary solutions.


Homework Equations


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The Attempt at a Solution


Using similar triange theroy, I get the following; x.y = 1.
Using pascals formula; 1^2 = (x + 1)^2 + (y + 1)6=^2
Solving the two equations gives an x value of -1.88 and a y value of -0.53.
Except I know I've made an mistake above and don't know how to continue. Can anyone help me please?
 
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  • #2
Hi Noir,

The length of the ladder is 10 m, so you should have x² + y² = 10². Use the fact that the ladder must touch the box to get another condition, and solve for x and y.

Hint: The second condition is (1/x) + (1/y) = 1.
 
  • #3
Thanks for the reply dx, sorry for my late response. Thanks for helping me but I have a quick question, how do you get the 1/x + 1/y = 1 relationship - It sounds stupid, and I know it works , but I seem to be missing a step in my head.
 
  • #4
The distance from the tip of the ladder on the x-axis to the edge of the box touching the x-axis is (x-1). The slope of the ladder (towards the y-axis) is y/x. Therefore, the height of the ladder at x = 1 is h = (x-1)(y/x). What must this height be if the ladder touches the box at this point?
 

FAQ: Solving a Complex Math Problem: Real and Imaginary Solutions

What is the difference between real and imaginary solutions in a complex math problem?

Real solutions are solutions that can be represented by a real number, while imaginary solutions are solutions that involve the use of the imaginary number "i". Real solutions are typically found when dealing with real-world problems, while imaginary solutions are often used in more abstract mathematical concepts.

How do you know if a complex math problem has real or imaginary solutions?

The presence of the imaginary number "i" in a mathematical expression is a good indication that the problem may have imaginary solutions. Additionally, solving the problem using algebraic techniques may reveal the presence of imaginary solutions.

What is the best approach to solving a complex math problem with both real and imaginary solutions?

The best approach is to first identify the type of problem you are dealing with and then use the appropriate techniques to solve it. For problems with both real and imaginary solutions, it may be necessary to use a combination of algebraic techniques and graphing methods to find all possible solutions.

Can a complex math problem have only real solutions?

Yes, it is possible for a complex math problem to have only real solutions. This often occurs in real-world scenarios where the problem can be represented by a real number, such as finding the distance between two points on a graph.

Are imaginary solutions relevant in real-life applications?

While imaginary solutions may not have a direct real-life application, they are still important in understanding and solving complex mathematical problems. Many real-world problems can be modeled and solved using complex numbers and their solutions involving imaginary numbers.

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