SOLUTION: What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Very confused.

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Question 139569: What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer.
Very confused.
Found 2 solutions by Edwin McCravy, solver91311:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. 


I will give all three cases and examples.  The answer to
your question is case 2 below, because  is the
same as saying  is negative.

---------------------------------------

The quadratic equation



has solution



The expression under the radical is called

the discriminant 

So that makes the solution



Since the value of  is under the radical

1. if D is positive, there are two different real solutions
2. if D is negative, there are two conjugate complex (imaginary) solutions
3. if D is zero, there is ONE solution with multiplicity 2.
-----------------------------------------
Example 1:  

Here , , 



Since D is a positive number, 24, then if
 were to be solved there would be
two different real solutions.
-----------------------------------------
Example 2:  

Here , , 



Since D is a negative number, -92, then if
 were to be solved there would be
two conjugate complex (imaginary) solutions.
------------------------------------------
Example 3:  

Here , , 



Therefore if  is solved there will be
ONE solution with multiplicity 2.

Edwin
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!
I presume by 'D' you mean the determinant in the quadratic equation. If D is less than zero, then you are faced with the problem of taking the square root of a negative number. If you think about that for a moment, you will realize it is an impossible situation. Taking a square root means finding two equal factors for a number and if the factors are equal, they must have the same sign. If two numbers with the same sign are multiplied together, you always get a positive result. So now what?

Mathematicians invented a special number, called an imaginary number, i, that is defined thusly: . Turns out that this is the only imaginary number needed because any negative number can be expressed as the product of -1 and the opposite of the number. -2 is the same as (-1)*2, for example. Since , if you had to deal with , you could change it to .

What happens when the discriminant is negative, or less than 0, is this:

where becomes

but we say that so:

(Note that if D < 0, -D > 0)

So D < 0 means that you will have a conjugate pair of complex roots of the form where and are real numbers and i is defined by Furthermore, solutions to the general quadratic with D < 0 will be where and

Note that complex roots ALWAYS come in conjugate pairs. If you find a complex root to a polynomial equation that is , then it is guaranteed that is also a root. This is also true for higher degree polynomial equations. It is no coincidence that quadratic (degree 2) equations have 2 roots. In fact, cubics (3rd degree) equations have 3 roots, quartics (4th degree) equations have 4 roots, and nth degree equations have n roots. As a consequence of the fact that complex roots must come in pairs, equations of odd degree (3rd, 5th, etc.) MUST have at least one real root.

Wait a minute, you say! Yes, you have solved quadratics where you only got one answer. In fact, you may have been told that if D=0 there is only 1 real root. To that I say, 'not so fast, mathematics breath'.

If D=0 then you have a perfect square trinomial, something like which factors to . [You can prove that by applying the quadratic equation to where , , and . D will come out to be ]

Just because the factors are identical doesn't mean there aren't 2 of them. And if there are 2 factors, there are also 2 roots. They just happen to be identical. Some mathematicians use the language 'one real root with a multiplicity of 2' Same thing as far as I'm concerned.
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