Polynominal question. Please help!!!!?

Question

What is the smallest positive integer value k can take if the equation has 2 RATIONAL solutions.
the eqn is: x^2 + (k-2)x - 4k = 0

Answer 1

find the discriminant

b^2 - 4ac

(k - 2)^2 - 4(1)(-4k)

This must equal 0 or be a perfect square to have rational roots.

so simplify and graph...

k^2 - 4k + 4 +16k

k^2 + 12k + 4

If k = 0, then this is 4.. which is a perfect square
but k must be greater than 0...

If you graph this function and check integer values, the first positive integer value that gives you a perfect square (that isn't negative) is 3 (which gives you 49)

So, K must be 3.

Answer 2

k^2 - 4k + 4 + 16k = n^2
k^2 + 12k + 4 = n^2
k^2 + 12k + 36 -36 + 4 = n^2
(k+6)^2 = n^2 + 32
k = - 6 + √(n^2 + 32)
when n = 2, k = 3

Answer 3

Hmm... well.... I would say graph it and experiment. Start with k equal to 1, so you'd be graphing x^2 - x - 4 = 0... keep moving up... what you're looking for is when the parabola's vertex hits the x-axis. Then you know there's only ONE rational solution; namely, the value of x where the vertex hits the axis... and that is one k-value higher than you need to go.

Sorry I can't give you a definitive answer; I am too lazy to go upstairs and get my calculator....