Question

Find two values of k so that each trinomial can be factored over the integers. 12x^2+kx+14

Answer 1

just, hint first.
D = m^2
wth D = discriminant and m integer

Answer 2

169 is works :)

Answer 3

12x^2+kx+14

D = m^2
k^2 - 4(12)(14) = m^2
k^2 - 672 = m^2
k^2 - m^2 = 672

case I :
(k+m)(k-m) = 336 * 2
k+m = 336
k-m = 2
--------- (+)
2k = 338
k = 169

k-m = 2
169-m=2
m=167

case II :
k^2 - m^2 = 672
(k+m)(k-m) = 168 * 4
k+m = 168
k-m=4
-------- (+)
2k = 172
k = 86

so, k = 86 is works too
for the others, maybe there are many possibles for k...

Answer 4

This way may be simpler.
The 2 values of k are : k = -26 and k = 26.
a) k = -26. This makes a + b + c = 0 (Tip 1). In this case, one real root is (1) and the other id (c/a = 14/12)
y = (x -1) (12x - 14)
b) k = 26. This make a - b + c = 0. One real root is (-1) and the other is (-c/a = -14/12)
y = (x + 1)(12x + 14)

REMINDER. Quick solving quadratic equations
a) Tip 1. When a + b + c = 1, one real root is (1) and the other is (c/a)
b) Tip 2. When a - b + c = 0, one real root is (-1) and the other is (-c/a).
Remember these 2 TIPs. They will save you lot of time.

Answer 5

good idea :)