SOLUTION: find the range of values of k for which the equation x^2-2kx+k^2-2k=6 has real roots. find the roots in terms of k

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Question 1092480: find the range of values of k for which the equation x^2-2kx+k^2-2k=6 has real roots. find the roots in terms of k
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

if the discriminant b%5E2+%96+4ac%3E0, the equation x%5E2-2kx%2Bk%5E2-2k=6 will have two real roots
if the discriminant b%5E2+%96+4ac=0, the equation x%5E2-2kx%2Bk%5E2-2k=6 will have one real root
if the discriminant b%5E2+%96+4ac%3C0, the equation x%5E2-2kx%2Bk%5E2-2k=6 will have no real roots
so, in your case we have
x%5E2-2kx%2Bk%5E2-2k=6
x%5E2-2kx%2B%28k%5E2-2k-6%29=0->a=1,b=2k,c=k%5E2-2k-6
b%5E2+%96+4ac=0
%282k%29%5E2+%96+4%2A1%28k%5E2-2k-6%29=0
4k%5E2+%96+4k%5E2%2B8k%2B24=0
8k%2B24=0
8k=-24
k=-24%2F8
k=-3-> for this value of k your equation will have one real solution
x%5E2-2%28-3%29x%2B%28-3%29%5E2-2%28-3%29-6=0
x%5E2%2B6x%2B9%2B6-6=0
x%5E2%2B6x%2B9=0
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2B6x%2B9%29+

and, first integer greater than -3 is -2
if k=-2 your equation will have two real solutions
x%5E2-2%28-2%29x%2B%28-2%29%5E2-2%28-2%29-6=0
x%5E2%2B4x%2B4%2B4-6=0
x%5E2%2B4x%2B2=0
+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+x%5E2%2B4x%2B2%29+