SOLUTION: Find the value(s) of k such that the equation f(x + k) = 0 has two equal real roots. f(x) = - (1/2)x^2 - 5x + 4 please show steps, thank you so much!

Click here to see ALL problems on Rational-functions

Question 252418: Find the value(s) of k such that the equation f(x + k) = 0 has two equal real roots.
f(x) = - (1/2)x^2 - 5x + 4

please show steps, thank you so much!
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Find the value(s) of k such that the equation f(x + k) = 0 has two equal real roots.
f(x) = - (1/2)x^2 - 5x + 4
---------------------------------
Find f(x+k):
f(x+k) = (1/2)(x+k)^2-5(x+k)+4
f(x+k) = (1/2)(x^2+2kx+k^2 - 5x-5k+4
f(x+k) = (1/2)(x^2+(2k-5)x + (k^2-5k+4)
f(x+k) = (1/2)x^2 + [(2k-5)/2]x + [k^2-5k+4]/2
----------------------------------------------------
You have a quadratic with
a = 1/2
b = (2k-5)/2
c = (k^2-5k+4)/2
-------------------------
Determine when f(x+k) will have two equal real roots.
That happens when the discriminant is zero
Solve:
b^2-4ac = 0
[(2k-5)/2]^2 - 4*(1/2)(k^2-5k+4)/2 = 0
[(1/4)(4k^2-20k+25)]-[k^2-5k+4] = 0
[k^2-5k+(25/4)]-[k^2-5k+4] = 0
9/4 = 0
---------
Well, I get a contradiction but I may have made
an arithmetic mistake somewhere along the line.
Check it out and see what you get.
===========================
Cheers,
Stan H.