Create a  problem situation in which only the positive solution is meaningful

First, you need to understand the question, right? You need to know what gives you more than one solution. The answer is an equation which contains a polynomial of at least degree 2. That means a quadratic equation or a cubic equation, or more, that has more than one solution. A typical example is a quadratic that has a positive root and a negative root: (x-a)(x+b)=0 or x^2+x(b-a)-ab=0, where a and -b are the roots.

Next we need a situation which implies that we need a positive solution only. An example is a geometrical figure like a triangle or rectangle which must, of course, have sides of positive length. Now we have to think of a situation which introduces the square of a number. In the case of geometrical figures this is likely to be the area.

Now we have to invent a problem. We start with a solution. Here's an example:

A rectangle has dimensions 9 by 4 so its area is 36 (never mind the dimensions at the moment). Then we create an unknown, x. Now we use x to define the dimensions of the rectangle. If we say that the length is 5 more than the width and the area is 36, then the solution is: let x=width then length=x+5, then the area is x(x+5)=36. So x^2+5x=36, or the quadratic equation: x^2+5x-36=0=(x-4)(x+9). The roots are 4 and -9, but we can reject -9. Now we have the basics of the problem, let's make it more interesting. First add dimensions. Let's use feet. Replace the rectangle with a familiar rectangular object: a box, swimming pool, a garden, etc. OK here's the question.

A swimming pool's length is 5 feet more than its width. What is its perimeter if the area of the base is 36 square feet?

We know how to find the width (4 feet) from the solution above, so we find the length (4+5=9 feet). The perimeter adds an extra to the problem, because it uses the length and the width. Perimeter=2(length+width)=26 feet.

answered Jul 3, 2016 by Top Rated User (486,860 points)