Question 831305
Q:
Find the value(s) of k for which the equation {{{3x^2-(k+1)x+k-2=0}}} has one real double root
-----------------------------------------------
A:
The quadratic equation {{{ax^2 + bx + c = 0}}},a ≠ 0,  has one real double root if the discriminant {{{b^2 - 4ac}}} is equal to zero.

In {{{3x^2-(k+1)x+k-2=0}}}, a = 3, b = -(k + 1), and c = k - 2
{{{b^2 - 4ac }}} = {{{(-1)^2(k + 1)^2 - (4)(3)(k - 2)}}}
             ={{{k^2 + 2k + 1 - 12k + 24}}}
             ={{{k^2 - 10k + 25}}}, equate to zero  
{{{k^2 - 10k + 25 = 0}}}
{{{(k - 5)^2 = 0}}}
Therefore, {{{highlight(k = 5)}}}