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) About Me  (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 D+%3C+0 is the
same as saying D is negative.

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The quadratic equation

Ax%5E2%2BBx%2BC=0

has solution

x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29+

The expression under the radical is called

the discriminant D=B%5E2-4AC

So that makes the solution

x+=+%28-B+%2B-+sqrt%28+D+%29%29%2F%282A%29+

Since the value of D 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.
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Example 1:  3x%5E2+%2B+7x+-+2=0

Here A=3, B=7, C=-2

D+=+B%5E2-4AC+=+%287%29%5E2-4%283%29%28-2%29+=+49-25+=+24

Since D is a positive number, 24, then if
3x%5E2+%2B+7x+-+2=0 were to be solved there would be
two different real solutions.
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Example 2:  -3x%5E2+%2B+4x+-+9=0

Here A=-3, B=4, C=-9

D+=+B%5E2-4AC+=+%284%29%5E2-4%28-3%29%28-9%29+=+16-108+=+-92

Since D is a negative number, -92, then if
-3x%5E2+%2B+4x+-+9=0 were to be solved there would be
two conjugate complex (imaginary) solutions.
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Example 3:  8x%5E2+-+8x+%2B+2=0

Here A=8, B=-8, C=2

D+=+B%5E2-4AC+=+%28-8%29%5E2-4%288%29%282%29+=+64-64+=+0

Therefore if 8x%5E2+-+8x+%2B+2=0 is solved there will be
ONE solution with multiplicity 2.

Edwin

Answer by solver91311(24713) About Me  (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: i%5E2=-1. 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 sqrt%28ab%29=sqrt%28a%29sqrt%28b%29, if you had to deal with sqrt%28-2%29, you could change it to sqrt%28-1%29sqrt%282%29.

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

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ where D=b%5E2-4ac%3C0 becomes

x+=+%28-b+%2B-+sqrt%28+%28-1%29%28-D%29+%29%29%2F%282%2Aa%29+ but we say that sqrt%28-1%29=i so:

x+=+%28-b+%2B-+i%2Asqrt%28%28-D%29+%29%29%2F%282%2Aa%29+ (Note that if D < 0, -D > 0)

So D < 0 means that you will have a conjugate pair of complex roots of the form alpha%2B-beta%2Ai where alpha and beta are real numbers and i is defined by i%5E2=-1 Furthermore, solutions to the general quadratic with D < 0 will be alpha%2B-beta%2Ai where alpha=-b%2F2a and beta=sqrt%28-D%29%2F2a

Note that complex roots ALWAYS come in conjugate pairs. If you find a complex root to a polynomial equation that is alpha%2Bbeta%2Ai, then it is guaranteed that alpha-beta%2Ai 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 x%5E2-4x%2B4=0 which factors to %28x-2%29%28x-2%29=0. [You can prove that by applying the quadratic equation to x%5E2%2B2px%2Bp%5E2 where a=1, b=2p, and c=p%5E2. D will come out to be 4p%5E2-4p%5E2]

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.