Highschool maths help?

Question

Could someone please explain how to do this....
Use the discriminate to prove that kx^2-(k+1)x+1=0 has rational roots if k is rational....
Thanks

Answer 1

an equation: ax^2 + bx + c has rational roots, with a, b and c being rational, if b^2 - 4ac is greater than or equal to zero

here b = -(k+1), a = k, and c = 1

therefore b^2-4ac = (k+1)^2 -4(k)1)
= k^2 + 2k + 1 -4k
= k^2 -2k + 1
= (k-1)^2

as i said, for an equational to has rational roots, its discriminate must be greater or equal to zero
therefore we want (k-1)^2 to be greater than or equal to 0, since (k-1)^2 is a square number, if must be greater than zero for all real values of k.
hence works for all rational values of k

Answer 2

b^2-4ac = (k+1)^2-4(k)(1) = (k-1)^2 ≥ 0, if k is rational. Therefore, the equation has rational root(s)

Answer 3

d = (k + 1)^2 - 4k
d = k^2 + 2k + 1 - 4k
d = k^2 - 2k + 1
d = (k - 1)^2
√d = k - 1
Therefore if k is rational, the roots are rational.

Answer 4

kx^2-(k+1)x+1=0
use this formula: b^2-4ac
(k+1)^2-4(k)(1)=0
k^2+2k+1-4k=0
k^2-2k+1=0
use scientific calculator to dissolve it
and you'll get : k=1

Answer 5

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

delta = (-(k+1))^2 - 4k=k^2-2k+1=(k-1)^2

x1 =( -(k+1) +sqrt (k-1)^2 ) / 2k = -1 / k
x2 = ( -(k+1) - sqrt(k-1)^2 ) / 2k = -1