SOLUTION: Find the possible values of d if real solutions exist for x² + 5x - 1 - d(x²+1) = 0

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Question 847477: Find the possible values of d if real solutions exist for x² + 5x - 1
- d(x²+1) = 0
Answer by swincher4391(1107) About Me  (Show Source):
You can put this solution on YOUR website!
After doing some distributive property you'll end up with this result.
-dx^2-d+x^2+5x-1 = 0
Let's group the x^2,x, and constant terms together like this:
(1-d)x^2 +5x +(-1-d)
From there we can satisfy our discriminant b%5E2-4ac+%3E=+0
5%5E2+-+4%281-d%29%28-1-d%29+%3E=0
25+%2B+4%281-d%5E2%29+%3E=+0
25+%2B+4+-+4d%5E2+%3E=0
29+%3E=+4d%5E2
4d%5E2%3C=+29
d%5E2+%3C=+29%2F4
d+%3C=+0+%2B-+sqrt%2829%2F4%29 <--- sorry I have to put that 0 in or the +- doesn't show
-sqrt%2829%29%2F2+%3C=+d+%3C=+sqrt%2829%29%2F2