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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?

 Jan 27, 2022
edited by Guest  Jan 27, 2022
 #1
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Yeah, I came across this problem too, can someone help me?

 Jan 27, 2022
 #2
avatar+36417 
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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 ?

 Jan 27, 2022
 #3
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I got ten, but it says that there are multiple solutions.

 Jan 27, 2022
 #4
avatar+36417 
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(-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
edited by ElectricPavlov  Jan 27, 2022
 #5
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its 10,5

 Feb 3, 2022

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