For what values of k does the equation kx^2+k=8x-2kx have 2 zeros?

Question

For what values of k does the equation kx^2 + k = 8x - 2kx have 2 zeros? Show your work.

Answer 1

Put it into standard form and factor or use the quadratic formula.

You want the discriminate to be positive
in standard form b=coeff of x: 2k-8
a = k
c = k

now plug and chug and find values of k that make the discriminant positive
b^2 - 4ac =

Answer 2

kx^2+k=8x-2kx kx^2 -8x +2kx +k = 0 kx^2 - x(8-2k) + k = 0 You right, two distinct roots will be then D>0 lets calculate it D = (8-2k)^2 - 4k*k 64 - 32k + 4k^2 -4k^2 > 0 -32k > -64 k < 2 Calculations is ok. "but this answer doesn't make sense because when i sub in a number greater than two into the equation, i still get a 2 roots" - so what do you want? You need 2 different roots, you get 2 different roots

Answer 3

kx^2 + k = 8x - 2kx
kx^2 + k - 8x - 2kx = 0
kx^2 - (8 + k)x + k = 0

in x = (-b ± sqrt(b^2 - 4ac))/(2a)

all you need is

b^2 - 4ac = 0

(-8 + k)^2 - 4(k)(k) = 0
(k - 8)^2 - 4k^2 = 0
((k - 8)(k - 8)) - 4k^2 = 0
(k^2 - 8k - 8k + 64) - 4k^2 = 0
k^2 - 16k + 64 - 4k^2 = 0
-3k^2 - 16k + 64 = 0
-(3k^2 + 16k - 64) = 0

k = (-b ± sqrt(b^2 - 4ac))/(2a)

k = (-16 ± sqrt(16^2 - 4(3)(-64)))/(2(3))
k = (-16 ± sqrt(256 + 768))/6
k = (-16 ± sqrt(1024))/6
k = (-16 ± 32)/6
k = (16/6) or (-48/6)
k = (8/3) or -8

ANS : (8/3) or -9