Solving Quadratic Equations
Date: 03/07/97 at 20:49:53 From: Caitlin Subject: Using algebra to find dimensions of a rectangle A certain rectangle has an area of 80 square units. Its length is one more than three times its width. What are the dimensions of the rectangle? Draw a diagram, solve the problem, and write an equation. I drew and labeled the diagram with the length as 3x+1 and the width as x. My equation is (3x+1)(x) = 80. I am stuck on how to solve the equation. I got as far as 3x^2 + x = 80. Now what do I do?
Date: 03/10/97 at 04:26:27 From: Doctor Mike Subject: Re: Using algebra to find dimensions of a rectangle Dear Caitlin, You made a very good start on the problem. My only suggestion so far is that "x" does not always have to be the unknown. If you had used "w" for width, and 3*w+1 for the length, then it's easy to keep track of what the unknown means when you get to the end of the problem. This is not an error, just something to think about when you do more and more complicated problems. If you subtract 80 from both sides you get 3x^2 + x - 80 = 0, which is a quadratic equation in standard form. This kind of problem comes up a lot and there are two main ways to solve it. 1. If you can factor the equation into the product of 2 things, then you can make good use of a well-known fact about numbers: If A*B = 0, then either A = 0 or B = 0 Your equation is sort of tough to factor. The factored version is (3x+16)*(x-5) = 0. Go ahead and multiply it out to see that it is the same. Now we know that the width x must satisfy EITHER 3x+16 = 0 OR x-5 = 0. The width has to be a positive number and only one of these solutions is greater than 0. (Keep in mind that the main reason anybody factors quadratic expressions is precisely because of that "A*B = 0" rule above.) 2. It's time you should memorize the quadratic formula. It says that *IF* an equation is in the form A*x^2+B*x+C = 0, then the 2 solutions for that equation (also called zeros because the right side is zero) are given by the formulas: -B + sqrt(B^2 -4*A*C) -B - sqrt(B^2 -4*A*C) x = ----------------------- x = ----------------------- 2*A 2*A You should go ahead and try this method also. It is very valuable because factoring can often be difficult. You will see again that one x-value solution is positive and one is negative. Only the positive one makes sense. By the way, "sqrt" means the square root function. Of course, you should make sure you get the same answer by method 1 and by method 2, and after you get your answer, go back and check to make sure that it works. In your problem that means to multiply the width x by the length 3*x+1 and verify that you really get 80 square units. I hope this helps. -Doctor Mike, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum