Question 1206552
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Discriminant formula
{{{d = b^2 - 4ac}}}


Rule: If {{{d > 0}}}, then the quadratic will have two distinct roots.


In this case,
a = k
b = -3+2k
c = 1


It might help to think of {{{kn^2 - 3n + 2kn + 1}}} as {{{kn^2 + (-3+2k)n + 1}}} so you can match those terms up to the template {{{an^2 + bn + c = 0}}}


Let's find the expression for the discriminant.
{{{d = b^2 - 4ac}}}


{{{d = (-3+2k)^2 - 4k(1)}}}


{{{d = (9-12k+4k^2) - 4k}}}


{{{d = 4k^2-16k+9}}}


Now consider the equation {{{4k^2 - 16k + 9 = 0}}}
Use the quadratic formula, which I'll leave the scratch work for the student to do, to find the following roots {{{k = (4-sqrt(7))/2 = 0.677124}}} and {{{k = (4+sqrt(7))/2 = 3.322876}}} Both of those decimal values are approximate.


If {{{k < (4-sqrt(7))/2}}} or {{{k > (4+sqrt(7))/2}}} then {{{4k^2 - 16k + 9 > 0}}} to lead back to {{{d > 0}}}. This will cause {{{kn^2 - 3n + 2kn + 1}}} to have two distinct roots for variable n.


If {{{k = (4-sqrt(7))/2}}} or {{{k = (4+sqrt(7))/2}}}, then {{{d = 0}}} and it will make {{{kn^2 - 3n + 2kn + 1}}} have exactly one real root for variable n.


If {{{(4-sqrt(7))/2 < k < (4+sqrt(7))/2}}} then it leads to {{{d < 0}}} and causes {{{kn^2 - 3n + 2kn + 1}}} to have no real roots. The two roots would instead be non-real complex numbers.
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