Question 962773
The quadratic formula is about a "general"
quadratic equation, namely:
{{{ f(x) = a*x^2 + b*x + c }}}
Then you set {{{ f(x) = 0 }}} to find the "roots"'
which are the solutions
{{{ a*x^2 + b*x + c = 0 }}}
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Suppose your equation for finding roots looks like:
{{{ -4x^2 + 6x - 11 = 0 }}}
Then, comparing to the "general" equation:
{{{ a = -4 }}}
{{{ b = 6 }}}
{{{ c = -11 }}}
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Now you have to plug these values into the "general"
formula for finding the roots:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
The tricky thing about this formula is that it is TWO formulas:
{{{x[1] = (-b + sqrt( b^2-4*a*c ))/(2*a) }}}
and
{{{x[2] = (-b - sqrt( b^2-4*a*c ))/(2*a) }}}
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That's because of the + or - in front of the square root sign
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Now plug in values:
{{{x = (-6 +- sqrt( 6^2 - 4*(-4)*(-11) ))/(2*(-4)) }}} 
{{{x = (-6 +- sqrt( 36 - 176 ))/(-8) }}} 
{{{ x = ( -6 +-sqrt( -140 )) / (-8) }}}
{{{ x[1] = ( -6 + 11.8322i ) / (-8) }}}
{{{ x[1] = 3/4 - 1.4790i }}}
That is one complex solution
The other one is:
{{{ x[2] = ( -6 - 11.8322i ) / (-8) }}}
{{{ x[2] = 3/4 + 1.4790i }}}
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So, you ended up with what they call a pair of
complex conjugate roots. That just means they
look like:
{{{ A + Bi }}} and {{{ A - Bi }}}
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Then there is the whole discussion of the "discriminant"
which is the {{{ b^2-4*a*c  }}} part of the formula
Because this is under the square root sign, the
"discriminant" will tell you whether you have:
(1) one "double" root in the case where:
{{{ b^2 = 4a*c }}}
(2) two rational roots when:
{{{ b^2 > 4a*c }}}
(3) that pair of complex conjugate roots where:
{{{ b^2 < 4a*c }}}
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The picture of the quadratic ( parabola ) tells you right
away which case you have:
(1) the parabola just touches the x-axis
(2) the parabola crosses the x-axis in 2 places
(3) the parabola never touches the x-axis
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Here's a plot of the 3 cases of the discriminant:
{{{ graph( 400, 400, -5, 5, -5, 5, 2x^2 + 4x + 2, 2x^2 + 3x + 2, 2x^2 + 5x + 2  ) }}} 
Hope this helps