x^2-2(k+1)x+k^2 solveroots are real and equal
Anonymous:
Not sure what's needed, but maybe this is helpful. If the roots are to be real and equal, then the discriminant needs to be equal to 0. So 4(k+1)^2 - 4k^2 = 0 => k = -1/2. With that value of k, the solution for x is x = 1/2.
you are right
dude
Glad to help. Shame there was no space left for an actual answer!
I will report answer
you give me answer
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Answer:
k = -1/2
x = 1/2
For the roots are to be real and equal, the discriminant must be equal to 0.
For the given quadratic
x² - 2(k+1)x + k²
the discriminant is
[ 2(k+1) ]² - 4(1)(k²) = 4(k+1)² - 4k² = 8k + 4.
So the roots are real and equal
=> discriminant is zero
=> 8k + 4 = 0
=> k = -1/2
Thus the quadratic must actually be
x² - 2( -1/2 + 1 )x + (-1/2)² = x² - x + 1/4
Solving for the roots:
x² - x + 1/4 = 0
=> 4x² - 4x + 1 = 0
=> ( 2x - 1 )² = 0
=> 2x - 1 = 0
=> x = 1/2
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