anonymous
  • anonymous
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)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
The 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): \[(-b \pm \sqrt{b ^{2}-4ac})/2a\] \[(-(-6) \pm \sqrt{6^{2}-4*1*4)}/2*1\] = \[6 \pm \sqrt{36-32}/2=6 \pm \sqrt {4}/2 = 3 \pm 1 = 4,2\]
anonymous
  • anonymous
And check my work via FOIL or whatever method you prefer.
anonymous
  • anonymous
Oh, and something is wrong since 4*2 does not equal 4.

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anonymous
  • anonymous
im 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
  • anonymous
Yep.
anonymous
  • anonymous
Like 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
  • anonymous
atleast it was caught XD

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