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Renee99

  • 3 years ago

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?

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  1. Mertsj
    • 3 years ago
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    A

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

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

  4. experimentX
    • 3 years ago
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    i go with campbell

  5. campbell_st
    • 3 years ago
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    the square root symbol

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

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

  8. Renee99
    • 3 years ago
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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.

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

  10. Mertsj
    • 3 years ago
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    You are correct.

  11. campbell_st
    • 3 years ago
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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

  12. Renee99
    • 3 years ago
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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??

  13. campbell_st
    • 3 years ago
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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

  14. Mertsj
    • 3 years ago
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    C is correct.

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

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

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

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

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

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

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

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

  23. Mertsj
    • 3 years ago
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    ok

  24. Renee99
    • 3 years ago
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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

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

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

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

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

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

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

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

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

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

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

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

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

  37. Mertsj
    • 3 years ago
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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

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

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

  40. Renee99
    • 3 years ago
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    Thank you!!

  41. Mertsj
    • 3 years ago
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    yw

  42. Renee99
    • 3 years ago
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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

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