SOLUTION: Find the range of k for which x^2-(k+2)x+k+5=0, has two different negative solutions Please show all working out. Thank you

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Question 1019225: Find the range of k for which x^2-(k+2)x+k+5=0, has two different negative solutions
Please show all working out. Thank you

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-%28k%2B2%29x%2Bk%2B5%22%22=%22%220

The solution using the quadratic formula:



simplifies to

%28k%2B2+%2B-+sqrt%28k%5E2-16%29%29%2F2

We only need to require that the larger
solution, the one with the plus sign before
the square root, be negative, so

%28k%2B2+%2B+sqrt%28k%5E2-16%29%29%2F2%3C0 

The discriminant must be positive to prevent
getting 1 single or 2 imaginary solutions, so 
setting the discriminant greater than 0,

k%5E2-16%3E0

That has solutions k < -4 or k > 4.
If k > 4 then certainly the larger solution 
will be positive.

So we must take k < -4

%28k%2B2+%2B+sqrt%28k%5E2-16%29%29%2F2%3C0

Multiplying both sides by 2

k%2B2+%2B+sqrt%28k%5E2-16%29%3C0

sqrt%28k%5E2-16%29%3C-k-2

k%5E2-16+%3Ck%5E2%2B4k%2B4

-16+%3C4k%2B4

-20+%3C4k

-5%3Ck

Therefore to have both solutions negative,

-5%3Ck%3C-4

Edwin