Question 139569
What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. 
<pre><font size = 4 color = "indigo"><b>

I will give all three cases and examples.  The answer to
your question is case 2 below, because {{{D < 0}}} is the
same as saying {{{D}}} is negative.

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The quadratic equation

{{{Ax^2+Bx+C=0}}}

has solution

{{{x = (-B +- sqrt( B^2-4AC ))/(2A) }}}

The expression under the radical is called

the discriminant {{{D=B^2-4AC}}}

So that makes the solution

{{{x = (-B +- sqrt( D ))/(2A) }}}

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^2 + 7x - 2=0}}}

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

{{{D = B^2-4AC = (7)^2-4(3)(-2) = 49-25 = 24}}}

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

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

{{{D = B^2-4AC = (4)^2-4(-3)(-9) = 16-108 = -92}}}

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

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

{{{D = B^2-4AC = (-8)^2-4(8)(2) = 64-64 = 0}}}

Therefore if {{{8x^2 - 8x + 2=0}}} is solved there will be
ONE solution with multiplicity 2.

Edwin</pre>