What is the sum of all possible values of for which x^2 + kx - 10x + 36 + 12x - 8 is the square of a binomial?
+128407 x^2 + kx -10x + 36 + 12x - 8 combine like terms
x^2 + (12 - 10 + k)x + 28
x^2 + ( 2 + k)x + 28
If this can be written as the square of a binomial the discrminant = 0
So
(2 + k)^2 - 4 (1) (28) = 0
(2 + k)^2 = 112
2 + k = ±√112 = ± 4√ 7
k = ± 4√7 - 2
The sum of all possible k's = (4√ 7 - 2) + ( -4 √ 7 -2 ) = -4
+36915 From CPhill answer :
x^2 + ( 2 + k)x + 28
the roots are the same R1 = R2
From Vieta
R1R1 = 28 so R1 = +- sqrt 28
and - ( R1 + R1) /a = -R1 - R1 = 2 + k (a = 1 in this case)
- 2R1 = 2+k
- 2(+- sqrt28) - 2 = k
- 4 sqrt7 -2 = k and 4 sqrt 28 -2 = k summed = -4
(as Chris found via different method)
