Question 1177949
Determine the exact value(s) of{{{ k}}} such that{{{ (k+2)x^2 -6 = kx -1}}} has one solution.

{{{(k+2)x^2 -6 - kx +1=0}}}

{{{(k+2)x^2- kx  -5 =0}}}

=> {{{a=(k+2)}}}, {{{b=-k}}},{{{ c=-5}}}

use discriminant:  

If discriminant,{{{ b^2 - 4ac > 0}}}, two solutions exist. 
If discriminant, {{{b^2 - 4ac = 0}}}, one solution exists. 
If discriminant, {{{b^2 - 4ac < 0}}}, no solution exists

so, you need:

{{{b^2-4ac=0 }}}....substitute {{{a}}},{{{b}}}, and {{{c}}}

{{{(-k)^2-4(k+2)(-5)=0 }}}

{{{k^2+20k+40=0}}} .........using quadratic formula, we get

{{{k = 2 (sqrt(15) - 5)}}}

{{{k = -2 (5 + sqrt(15))}}}


{{{ (2(sqrt(15) - 5)+2)x^2 -6 = 2 (sqrt(15) - 5)x -1}}}

{{{(2sqrt(15) - 8)x^2 - 6=2 (sqrt(15) - 5)x -1}}} -> has one solution

or
{{{ (-2(sqrt(15) +5)+2)x^2 -6 =-2(sqrt(15) + 5)x -1}}}
{{{-(2sqrt(15)+8)x^2 - 6=-2(sqrt(15) + 5)x -1}}}-> has one solution