SOLUTION: A quadratic function h has the rule h(x) = x2 - 4x +5. Find the value of k for which h(x) +k =0 has two positive solutions.

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Question 1028569: A quadratic function h has the rule h(x) = x2 - 4x +5. Find the value of k for which h(x) +k =0 has two positive solutions.
Found 2 solutions by Fombitz, rothauserc:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-4x%2B5%2Bk=0
%28x%5E2-4x%2B4%29%2B5%2Bk=4
%28x-2%29%5E2=-1-k
%28x-2%29%5E2=-%281%2Bk%29
x-2=0+%2B-+sqrt%28-1-k%29
x=2+%2B-+sqrt%28-1-k%29
So use the lower zero,
2-sqrt%28-1-k%29%3E0
-sqrt%28-1-k%29%3E-2
sqrt%28-1-k%29%3C2
-1-k%3C4
-k%3C5
k%3E-5

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
we are give
:
h(x) + k = 0
:
x^2 -4x +5 +k = 0
:
if k = -2, then
:
x^2 -4x +3 = 0
:
(x-3) * (x-1) = 0
:
x = 3 and x = 1
: