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anonymous

  • 5 years ago

How many real solutions are there to the equation shown below? x2 + 2x + 6 = 0

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  1. anonymous
    • 5 years ago
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    to know, you'll have to factor ^_^

  2. amistre64
    • 5 years ago
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    none: the discriminate is negative: 4 -24 = -20

  3. anonymous
    • 5 years ago
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    how come? .-.

  4. anonymous
    • 5 years ago
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    for a quadratic equation ax^2 + bx+ c Disc = b^2 -4ac if D<0 roots are imaginary

  5. amistre64
    • 5 years ago
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    that (b^2 -4ac) is negative

  6. anonymous
    • 5 years ago
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  7. anonymous
    • 5 years ago
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    i qot this we i factored it .

  8. amistre64
    • 5 years ago
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    right, the solutions are imaginary ;)

  9. anonymous
    • 5 years ago
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    i was thinking it Was 2 real solutions .

  10. amistre64
    • 5 years ago
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    if its got an "i" in it; that means "i"maginary. Its a cruel joke really, such a useful complex number and all, but to be dubbed imaginary all your life is kinda a misnomer

  11. radar
    • 5 years ago
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    \[x=-1\pm \sqrt{-5}\]

  12. amistre64
    • 5 years ago
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    it simply means it exists, but not in the realm of the "real" numbers. So it tends to not get invited to all the social gatherings and all the other numbers that hang out with it soon leave ...

  13. anonymous
    • 5 years ago
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    oh, I missed that ^_^"

  14. anonymous
    • 5 years ago
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    so i quess It foold Mee uqqh .

  15. anonymous
    • 5 years ago
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    \[ax^2 + bx + c = 0 \] \[\implies x=\frac{-b\pm \sqrt{b^2-4(a)(c)}}{2(a)}\] \[b^2 - 4(a)(c) < 0 \implies \text{No real solutions}\] Because the thing under the square root would be negative. \[b^2 - 4(a)(c) = 0 \implies \text{One real solution}\] Because the \(\pm\) bit would be \(\pm\) 0. \[b^2 - 4(a)(c) > 0 \implies \text{Two real solutions}\]

  16. amistre64
    • 5 years ago
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    in math, you have to use maths definitions...not english definitions lol

  17. amistre64
    • 5 years ago
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    "real" in math means: not having an "i"

  18. anonymous
    • 5 years ago
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    idk , I dont qet it Maybe one solution

  19. radar
    • 5 years ago
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    \[x=-1\pm \sqrt{5 \times i ^{2}}=-1\pm i \sqrt{5}\]

  20. amistre64
    • 5 years ago
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    No SOLUTIONS

  21. amistre64
    • 5 years ago
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    no REAL solutions lol

  22. anonymous
    • 5 years ago
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    If the part under the square root is negative you have no real soltions.

  23. anonymous
    • 5 years ago
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    but how do yu find if theres Some solutions or not how yu determine

  24. anonymous
    • 5 years ago
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    Look at the quadratic equation.

  25. anonymous
    • 5 years ago
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    Have you learned the quadratic equation?

  26. anonymous
    • 5 years ago
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    yea

  27. amistre64
    • 5 years ago
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    Home sweet home:

  28. anonymous
    • 5 years ago
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  29. anonymous
    • 5 years ago
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    Ok. So when you plug in a, b, and c into the quadratic equation. You see the part with the square root?

  30. anonymous
    • 5 years ago
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    What is the domain of the function F(x) graphed below?

  31. anonymous
    • 5 years ago
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    Ohh okayy i qet tht now .

  32. amistre64
    • 5 years ago
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    the set of all Real numbers

  33. anonymous
    • 5 years ago
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    The domain of the function is all the values you can give x and get real solutions for F(x). Since there are no vertical asymptotes or holes in the graph, the domain is \((-\infty,\infty)\)

  34. anonymous
    • 5 years ago
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    amistre64 Wats tht yu wrote the set of all real numbers

  35. anonymous
    • 5 years ago
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    -1 < x < 1

  36. amistre64
    • 5 years ago
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    Yes; the domain of your graph there is the "set of all Real numbers". All the numbers that live on the number line are in the domain

  37. amistre64
    • 5 years ago
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    the domain for f(x) = (x-1)^2 -1 is; any and all real numbers. The graph extends forever to the left, and forever to the right with no restrictions

  38. anonymous
    • 5 years ago
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    What is the y-coordinate of the vertex of a parabola with the following equation? y = x2 - 8x + 18

  39. amistre64
    • 5 years ago
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    (-b/2a, f(-b/2a) :)

  40. amistre64
    • 5 years ago
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    x=4; y = 4^2 -8(4) + 18

  41. amistre64
    • 5 years ago
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    16-32+18 = -16+18 = 2 (4,2)

  42. anonymous
    • 5 years ago
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    itss 18 right

  43. amistre64
    • 5 years ago
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    18 is the "y intercept" that is not the vertex for this graph.

  44. amistre64
    • 5 years ago
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    the vertex for this graph is (4,2) so the y coordinate of the vertex is 2

  45. amistre64
    • 5 years ago
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    the vertex is where the graph bends and starts going back around; its kinda like the "handle" of the graph. we can use it to move the graph anywhere we want of the graph paper

  46. anonymous
    • 5 years ago
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    ohh Okay but I thought it was 18 because i factor it & i thouqht they was asking for tha y intercept But read it to fast & put tha wrong answer . dummie Mee

  47. amistre64
    • 5 years ago
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    lol... I do that alot as well; I read what I want to see in it and get a wrong answer ;)

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