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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)

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  1. anonymous
    • 5 years ago
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    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): \[(-b \pm \sqrt{b ^{2}-4ac})/2a\] \[(-(-6) \pm \sqrt{6^{2}-4*1*4)}/2*1\] = \[6 \pm \sqrt{36-32}/2=6 \pm \sqrt {4}/2 = 3 \pm 1 = 4,2\]

  2. anonymous
    • 5 years ago
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    And check my work via FOIL or whatever method you prefer.

  3. anonymous
    • 5 years ago
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    Oh, and something is wrong since 4*2 does not equal 4.

  4. anonymous
    • 5 years ago
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    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 -?

  5. anonymous
    • 5 years ago
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    Yep.

  6. anonymous
    • 5 years ago
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    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.

  7. anonymous
    • 5 years ago
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    atleast it was caught XD

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