Find all real numbers k for which the equation (k-5)x^2-kx+5=0 has exactly one real solution. I don't understand how to start or answer this problem. I don't know how to find the discriminant, can someone help me?

Guest Jan 27, 2022

edited by
Guest
Jan 27, 2022

#2**+1 **

The 'discriminant' is the b^2 - 4ac portion of the Quadratic Formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a = k-5 b = -k c = 5

When the discriminat = 0 there is only one solution to the quadratic

(-k)^2 - 4 (k-5)(5) = 0 <======== can you take it from here....solve for k ?

ElectricPavlov Jan 27, 2022

#3

#4**+1 **

(-k)^2 - 4 (k-5)(5) = 0

k^2 - 20k + +100 = 0

(k-10)(k-10) = 0 k= 10 or 10 (so really, only ONE solution)

Are they ALSO asking for the solution to the ORIGINAL equation when k = 10 maybe ?????

ElectricPavlov
Jan 27, 2022