Integers quadratic equations

When you set 3x = 0, you get that x = 0. When x + 4 = 0, you get that x = -4. Neither of these numbers works in the original problem, because you need a positive integer, and neither is positive. This problem has no answer. Even though the quadratic equation has solutions, neither works with the question. 

Write the two consecutive integers as n and n + 1. Multiply them together and set the equation equal to 20. You get a quadratic equation that s solved by setting the equation equal to 0, factoring, and solving for the numbers. 

The quadratic equation gives you two different answers. When n = 4, you get the two consecutive integers 4 and 5. The product of 4 and 5 is, indeed, 20. But what about the solution n = -5 If n = -5, then n+l=-5+l = -4. The product of -5 and -4 is also 20. This problem has two different solutions. As long as you re happy with negative integers, too, then you accept both sets of answers. 

The product of the two consecutive odd integers is written as rt(n + 2). Set that product equal to 143, set the equation equal to 0, and solve the quadratic equation that results. 

In high school algebra you frequently used factoring to solve quadratic equations. However, in this and virtually all other real problems you will encounter in chemistry, neither the numerical coefficients nor the solutions will turn out to be integers, so factoring is not useful. The general solutions to the equation ax2 + bx + c = 0 are given by the expression... 

The only solutions, L(x), of this equation that are quadratically integrable are those for which the coefficient of L is zero or a positive integer this condition requires that the parameter n be an integer such that n — (I 1) > 0 orn > I 1. Since the least value of / is zero we have the quantization conditions... 

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