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Renee99

How can you tell when a quadratic equation has no real solutions? A. when the radicand is negative B. when b in the quadratic formula is greater than the radicand C. when the radicand equals zero D. when the radicand is not a perfect square I believe the answer is A? is that right?

  • 2 years ago
  • 2 years ago

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  1. Mertsj
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    A

    • 2 years ago
  2. campbell_st
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    its A using the discriminant \[b^2 - 4ac\] < 0 no real solutions the radical will be negative

    • 2 years ago
  3. experimentX
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    whats radicand anyway??

    • 2 years ago
  4. experimentX
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    i go with campbell

    • 2 years ago
  5. campbell_st
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    the square root symbol

    • 2 years ago
  6. Mertsj
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    b^2-4ac

    • 2 years ago
  7. Mertsj
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    \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

    • 2 years ago
  8. Renee99
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    Thanks!! :) I also have this one... For which value of x does the graph of y = 2x2 − 7x + 6 cross the x-axis? A. −3/2 B. −2/3 C. 2 D. 3 I think its C.

    • 2 years ago
  9. Mertsj
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    \[b^2-4ac\] is the discriminant

    • 2 years ago
  10. Mertsj
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    You are correct.

    • 2 years ago
  11. campbell_st
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    you can substitute each value to find y = 0 or factorise (2x - 3)(x - 2) = 0 x = 3/2 and 2

    • 2 years ago
  12. Renee99
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    Thanks so much, here's another one i'm stuck on... What are the approximate solutions of 4x2 + 3 = −12x to the nearest hundredth? A. x ≈ −3.23 and x ≈ 0.23 B. x ≈ −2.72 and x ≈ −0.28 C. x ≈ 0.28 and x ≈ 2.72 D. x ≈ −0.23 and x ≈ 3.23 I think its C, but I'm not sure??

    • 2 years ago
  13. campbell_st
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    put it in standard form 4x^2 + 12x + 3 = 0 use the general quadratic formula \[x = (-b \pm \sqrt{b^2 - 4ac})/2a\] in your question a = 4, b = 12 and c = 3

    • 2 years ago
  14. Mertsj
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    C is correct.

    • 2 years ago
  15. Renee99
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    thanks so much!! :)

    • 2 years ago
  16. Mertsj
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    No. It's B

    • 2 years ago
  17. Mertsj
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    Both roots are negative.

    • 2 years ago
  18. Renee99
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    B? Really? are you sure?

    • 2 years ago
  19. Renee99
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    Okay, i'll trust you! :)

    • 2 years ago
  20. Mertsj
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    \[x=\frac{-12\pm \sqrt{96}}{8}\]

    • 2 years ago
  21. Renee99
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    I have one more question that's really confusing me, would you mind sticking around for one more?

    • 2 years ago
  22. Mertsj
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    \[x=\frac{-12+9.80}{8} or x=\frac{-12-9.80}{8}\]

    • 2 years ago
  23. Mertsj
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    ok

    • 2 years ago
  24. Renee99
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    *note* don't ask me why the numbers are spreed out like that, I just copied and pasted it onto here, that how it looks on my paper. Which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring? −b b2 − 4ac 2a Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 2x2 + 7x + 3 = 0

    • 2 years ago
  25. Mertsj
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    b^2-4ac

    • 2 years ago
  26. Renee99
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    is that the answer?

    • 2 years ago
  27. Mertsj
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    Well which part would you choose? -b? 2a????

    • 2 years ago
  28. Mertsj
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    \[b^2-4ac=7^2-4(2)(3)=49-24=25\]

    • 2 years ago
  29. Renee99
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    I don't know... This is really confusing me for some reason...

    • 2 years ago
  30. Renee99
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    so the answer is b^2-4ac?

    • 2 years ago
  31. Renee99
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    I'm still confused... :/

    • 2 years ago
  32. Renee99
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    or is the answer this: b² - 4ac = (7)² - 4(2)(3) = 49 - 24 = 25.

    • 2 years ago
  33. Mertsj
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    The discriminant, which is b^2-4ac, is the part of the quadratic formula which tells you about the roots.

    • 2 years ago
  34. Mertsj
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    Do you understand that the question has two parts? READ THE QUESTION!!!

    • 2 years ago
  35. Mertsj
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    PART 1: Which part of the quadratic formula tells you whether the quadratic equation can be solved by factoring?

    • 2 years ago
  36. Mertsj
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    The answer to that is :b^2-4ac

    • 2 years ago
  37. Mertsj
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    PART 2: Use the part of the quadratic formula that you chose above and find its value given the following quadratic equation: 2x2 + 7x + 3 = 0

    • 2 years ago
  38. Mertsj
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    The answer to that is 25

    • 2 years ago
  39. Renee99
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    ohhh okay, I understand now

    • 2 years ago
  40. Renee99
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    Thank you!!

    • 2 years ago
  41. Mertsj
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    yw

    • 2 years ago
  42. Renee99
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    I just wasn't looking at the problem correctly, sorry for the confusion and thanks again for breaking it down for me

    • 2 years ago
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