Question 201614
1. How to determine the number of solutions.
Let's look at the quadratic formula:
{{{x = (-b +- sqrt(b^2 - 4ac))/2a}}}
The key is the expression in the square root: {{{b^2 - 4ac}}}. In general there are three cases:<ol><li>{{{b^2 -4ac}}} is positive. The square root of a positive number is also some positive number. So in the numerator of the quadratic formula we will get two values: (-b + the square root) and (-b - the square root). So when {{{b^2 - 4ac > 0}}} we get two solutions.</li><li>{{{b^2 -4ac}}} is zero. The square root of zero is zero. So in the numerator we get (-b + 0) and (-b - 0). But both of these are equal to -b! So when {{{b^2 - 4ac = 0}}} we only get one solution.</li><li>{{{b^2 -4ac}}} is negative. And what is the square root of a negative number? What can we square and get a negative number as an answer? Answer: Nothing. You cannot square any Real number and get a negative. So when {{{b^2 -4ac < 0}}} there are no solutions.</li></ol>
2. Finding the equation from the solution(s)
One way to find solutions from the equation is to factor it. For example, solving
{{{x^2-5x+6=0}}}
we factor it:
{{{(x-2)(x-3)=0}}}
For a product to be zero one of the factors must be zero. In other "words":
x-2 = 0 or x-3 =0
Solving these we get:
x=2 or x=3
Now what you want is to be able to do this in reverse. Well all the steps above are reversible. Therefore, if we have two solutions: {{{x = x[1]}}} and {{{x = x[2]}}}, the equation is going to be:
{{{(x - x[1])(x - x[2]) = 0}}}
Some examples:
Solution: x = 1 or x = 6
Equation: (x-1)(x-6) = 0 which gives {{{x^2 -7x + 6 = 0}}}
Solution: x = 10 or x = 0
Equation: (x-10)(x-0) = 0 or (x-10)(x) = 0 which gives {{{x^2 -10x = 0}}}
Solution: x = -3 or x = 1.7
Equation: (x-(-3))(x-1.7) = 0 or (x+3)(x-1.7)=0 (You multiply out this one!?)
If you are given that there is only one solution to a quadratic equation then the equation is of the form: {{{(x - x[1])^2 = 0}}}. For exmaple, if the only solution to to a quadratic equation is 20, then the equation would be: {{{(x - 20)^2 = 0}}} which gives {{{x^2 -40x + 400 = 0}}}.<br>
3. Can different quadratic equations have the same solution? Well it depends on what you mean by "different". Yes. For example: {{{(x-4)(x-5) = 0}}} {{{2(x-4)(x-5) = 0}}} and {{{-5(x-4)(x-5) = 0}}} all have same solution: x-4 or x=5. But are these "different" equations? They sure look different. But if you divide both sides of the second equation by 2 you get the first equation. If you divide both sides of the third equation by -5 you get the first equation. So are these equations "different"? In my <i>opinion</i> these equations are not different and that the answer to question #3 is no.