Question 825876
{{{ 2x^2 - 5x - k = 0 }}}
Use the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 2 }}}
{{{ b = -5 }}}
{{{ c = -k }}}
the roots are imaginary when {{{ b^2 - 4*a*c }}}
is negative. Since {{{ b^2 }}} is always positive.
{{{ 4*a*c > b^2 }}}
{{{ 4*2*(-k) > (-5)^2 }}}
{{{ -8*k > 25 }}}
{{{ k < -25/8 }}} answer
( note that I had to reverse the inequality sign
when I divided by a negative number )
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Suppose {{{ k = -25/8 }}}
 {{{ b^2 - 4*a*c = (-5)^2 - 4*2*(-k) }}}
 {{{ b^2 - 4*a*c = 25 - 4*2*(-(-25/8)) }}}
 {{{ b^2 - 4*a*c = 25 - 25 }}}
{{{ b^2 - 4*a*c = 0 }}}
That means there is 1 real root
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Suppose {{{ k = -26/8 }}} or
{{{ k = -13/4 }}}
{{{ b^2 - 4*a*c = (-5)^2 - 4*2*(-k) }}}
{{{ b^2 - 4*a*c = 25 - 4*2*(-(-13/4)) }}}
{{{ b^2 - 4*a*c = 25 - 26 }}}
{{{ b^2 - 4*a*c = -1 }}}
That means there are 2 imaginary roots