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anonymous
 5 years ago
3 Parts to this question
a. Find the value of the discriminant
b. Give the nukber of real solutions
c. Find the real solutions, rounded to the nearest hundred.
r^(2) + 6r + 4 = 0
(this is a reallu weird question so if you dont understand it i understand)
anonymous
 5 years ago
3 Parts to this question a. Find the value of the discriminant b. Give the nukber of real solutions c. Find the real solutions, rounded to the nearest hundred. r^(2) + 6r + 4 = 0 (this is a reallu weird question so if you dont understand it i understand)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The discriminant = \[b ^{2}4ac = 6^{2}4*1*4 = 36  32 = 4\] # of Real Solutions: 2 Real solutions using quadratic formula (check my algebra carefully cause I tend to get sloppy): \[(b \pm \sqrt{b ^{2}4ac})/2a\] \[((6) \pm \sqrt{6^{2}4*1*4)}/2*1\] = \[6 \pm \sqrt{3632}/2=6 \pm \sqrt {4}/2 = 3 \pm 1 = 4,2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0And check my work via FOIL or whatever method you prefer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh, and something is wrong since 4*2 does not equal 4.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im confused how did ur 6 become (6) when it was positive to begin with shouldnt it have become (6) and htere for 6 + or  not 6 + or ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like I said I get sloppy with my algebra and arithmetic. Sorry. I think something is still wrong in there somewhere but I'm going to be lazy with it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0atleast it was caught XD
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