Question 1078340: X^2-2kx-8k^2
Find in terms of k, the least value for c for which the equation f(x)=c has a real solution for x.
Wth is this?
Thanks for the help
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! f(x) = c = x^2-2kx-8k^2
Subtract c from both sides:
x^2-2kx-8k^2-c = 0
This is a quadratic equation with coefficients 1, -2k and -8k^2 - c
In order for there to be one real solution b^2 - 4ac = 0
So we have (-2k)^2 - 4*(-8k^2 - c) = 0
4k^2 + 32k^2 + 4c = 0
9k^2 + c = 0
Or c = -9k^2
Thus c must be at least -9k^2 for one real solution
With this value for c, the equation becomes x^2-2kx-8k^2 + 9k^2 = 0 -> x^2-2kx + k^2 = 0
This can be factored as (x-k)(x-k) = 0, which gives one real solution, x = k.
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