SOLUTION: How do you know if a quadratic equation will have one, two, or no solutions? How do you find a quadratic equation if you are only given the solution? Is it possible to have differe
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Question 201614: How do you know if a quadratic equation will have one, two, or no solutions? How do you find a quadratic equation if you are only given the solution? Is it possible to have different quadratic equations with the same solution? Explain. Provide your classmate’s with one or two solutions with which they must create a quadratic equation
Found 2 solutions by jsmallt9, solver91311:
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
1. How to determine the number of solutions.
Let's look at the quadratic formula:
The key is the expression in the square root: . In general there are three cases:- 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 we get two solutions.
- 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 we only get one solution.
- 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 there are no solutions.
2. Finding the equation from the solution(s)
One way to find solutions from the equation is to factor it. For example, solving
we factor it:
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: and , the equation is going to be:
Some examples:
Solution: x = 1 or x = 6
Equation: (x-1)(x-6) = 0 which gives
Solution: x = 10 or x = 0
Equation: (x-10)(x-0) = 0 or (x-10)(x) = 0 which gives
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: . For exmaple, if the only solution to to a quadratic equation is 20, then the equation would be: which gives .
3. Can different quadratic equations have the same solution? Well it depends on what you mean by "different". Yes. For example: and 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 opinion these equations are not different and that the answer to question #3 is no.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
How do you know if a quadratic equation will have one, two, or no solutions?.
The short answer is, it is easy: They all have two solutions. The Fundamental Theorem of Algebra guarantees it. Now if what you really meant to ask is: How do you know if a quadratic equation will have one, two, or no solutions over the real numbers, then read on.
A quadratic equation of the form:
has a general solution:
The expression under the radical in the general solution, namely 
is called the discriminant. The discriminant can be evaluated to determine the character of the solutions of a quadratic equation, thus:
If 
, then
if 
, then the quadratic has two distinct real number roots. Furthermore, if 
is a perfect square number, then the roots will be rational, otherwise the roots of the equation will be a conjugate pair of irrational numbers of the form 
where 
if 
, then the quadratic has a single real number root with a multiplicity of 2. In this case the quadratic is a perfect square having two factors: (x - \alpha))
, hence 
is the root, and the multiplicity of 2 comes from the fact that there are two identical factors.
if 
, then the quadratic has no real number solutions. It has a conjugate pair of complex roots of the form 
where 
and 
is the imaginary number defined by 
======================================================
How do you find a quadratic equation if you are only given the solution?
Note, that should read "solutions"
If the solutions of a quadratic equation are and 
, then you can say:
and
, therefore
(x - \beta) = 0)
and finally,
x + \alpha\beta = 0)
Hence deriving the quadratic equation from the solutions. Note that if you are only given one number as a solution, then you have three possibilities. The first is that you may have a rational number, in which case you have two identical factors that become a perfect square trinomial: (x - \alpha) = 0 \ \ \Rightarrow\ \ x^2 - 2\alpha x + \alpha^2 = 0)
. The second possibility is that you are given an irrational number, such as 
with the possibility that 
. In this case, the second root is the conjugate of the given root, namely 
. The third possibility is that you are given a complex number root, 
. Again, the second root is the conjugate: 
======================================================
Is it possible to have different quadratic equations with the same solution?
Well, that depends on what you mean by 'different'. For a given 
, 
, and 
, 
is equivalent to, and has the same roots as 
for all 
(actually it is true for all 
but let's stick with rational coefficients for your purposes). Since the equations are equivalent, it is a stretch, for me anyway, to call them 'different.' However, what is true is that the analogous functions, i.e.  = k_iax^2 + k_ibx + k_ic)
are very different in that they all have different graphs.
So while it is true that  = x^2 - 6x + 5)
is different than  = 2x^2 - 12x + 10)
(see graphs below), the equations 
and 
are equivalent because you can multiply the second one by 
to obtain the first. Note that the graphs intersect the 
-axis at the same points:
======================================================
There should be enough information above for you to come up with a set of solutions for a quadratic for your classmates, particularly in the answer to the second question.
John
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