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anonymous

  • one year ago

I need to find the zeroes of f(x) = 5x2 + 2x + 1

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  1. anonymous
    • one year ago
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    @AliceCullen @ramseysa @Elsa213 @DaBest21 @thomaster

  2. BloomLocke367
    • one year ago
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    zeroes are just your solutions. It's where x=0, or where the graph crosses the x-axis. Do you know how to solve quadratics?

  3. anonymous
    • one year ago
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    No @BloomLocke367

  4. BloomLocke367
    • one year ago
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    Here, take a look at this tutorial to see if you understand. If you still need help, please do tag me and I'll try my best to help you! http://openstudy.com/users/bloomlocke367#/updates/55acffcee4b071e6530c96ce

  5. anonymous
    • one year ago
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    Wait, is -1/5 and 4/5

  6. anonymous
    • one year ago
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    Are those the zeroes?

  7. BloomLocke367
    • one year ago
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    Actually, I'm pretty sure the zeroes are complex roots. One moment. I have to double check my work.

  8. anonymous
    • one year ago
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    - 1/5 and - 4/5

  9. BloomLocke367
    • one year ago
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    No, those are not the roots. They are, in fact, complex. Do you know what a complex number is?

  10. anonymous
    • one year ago
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    No

  11. anonymous
    • one year ago
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    - 1/5 and - 2i/5

  12. BloomLocke367
    • one year ago
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    Basically, a complex number is a number that has an imaginary number in it. The often have a coefficient, i, and a constant. Here's an example of what it would look like: 3i+8. That's just an example, not the answer. Do you understand imaginary numbers? Where did you get those answers?

  13. anonymous
    • one year ago
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    Mathway, I suck at math and despise it. I'm sorry, I'm a history and English guy.

  14. anonymous
    • one year ago
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    Actually, I got them from Wolfram/Alpha

  15. BloomLocke367
    • one year ago
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    Really? huh. Try using the quadratic formula, which is \(\LARGE x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

  16. BloomLocke367
    • one year ago
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    In this case, a is 5, b is 2, and c is 1.

  17. anonymous
    • one year ago
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    (-2±√(2^2-4*5*1))/(2*5)

  18. anonymous
    • one year ago
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    Is it 1/10(-2 ± 4i)?

  19. anonymous
    • one year ago
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    @badmood

  20. radar
    • one year ago
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    Actually you were right. You stated you had to find the 0"s of f(x) = 5x^2 + 2x + 1 This is the value where f(x) = 0. Obviously f(x) depends on x and x can be an infinite number of values both positive and negative and complex. When x = 0, in this given function f(x) = 1, when x = -1, f(x) = 4 etc. for f(x) = 0 you would solve for x, and the quadratic equation would be a reasonable method, however when you graph this function you will find that the f(x) nevers gets to 0. This does not mean that the function does not exists it simply means there is nor real "0" for f(x) for any real value of x.

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