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EmmaH

  • 2 years ago

cansomeone show me how to solve this? Use the quadratic formula to find the zeros of the function. y = 16x2 + 40x + 25 {-1.25, 1.25} {-1.25} {-4, 5} {-0.8

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  1. amistre64
    • 2 years ago
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    do you recall the setup of the quadratic formula?

  2. EmmaH
    • 2 years ago
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    \[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

  3. amistre64
    • 2 years ago
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    good, and the general form of a quadratic is: \[ax^2 + bx + c\] pull the numbers from the problem that relate to the abc parts and what do you get worked out?

  4. EmmaH
    • 2 years ago
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    a=16 b=40 and 25=c

  5. amistre64
    • 2 years ago
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    good, i notice that there are no - signs to worry about so that looks good so far

  6. EmmaH
    • 2 years ago
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    what do i do next?

  7. amistre64
    • 2 years ago
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    plug in those numbers into their abc parts in the formula, the rest is just math for maths sake

  8. amistre64
    • 2 years ago
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    \[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\] \[x=\frac{ -40 \pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\]

  9. EmmaH
    • 2 years ago
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    \[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] is that right?

  10. amistre64
    • 2 years ago
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    looks good to me :)

  11. EmmaH
    • 2 years ago
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    ok then i get 40 pm 16-96/32?

  12. EmmaH
    • 2 years ago
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    but i would end up eith a negative number....

  13. amistre64
    • 2 years ago
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    \[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{4^2(10)^2-4(4^2)(25)} }{ 2(16) }\] \[\frac{ -40\pm 4\sqrt{10^2-4(25)} }{ 2(16) }\] \[\frac{ -40\pm 4\sqrt{25(4)-4(25)} }{ 2(16) }\] \[\frac{ -40}{ 2(16) }\] \[\frac{ -40}{ 2(4.4) }\] \[\frac{ -10}{ 2(4) }\] \[\frac{ -5}{ 4 }\]

  14. EmmaH
    • 2 years ago
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    you lost me at the second one you did how did you get all the the ^2?

  15. amistre64
    • 2 years ago
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    40^2 = (4*10)^2 = 4*10 * 4*10 = 4^2 * 10^2 but you dont have to do exactly what i did

  16. EmmaH
    • 2 years ago
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    uuuuh.....im sorry you lost me all i know is im suppose to square 40 is that what you were doin?

  17. EmmaH
    • 2 years ago
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    @amistre64 ?

  18. amistre64
    • 2 years ago
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    i was doing some manipulations so that i didnt have to worry about computing large numbers and such; i simplified it down to where i recognized the sqrt part goes to zero.

  19. amistre64
    • 2 years ago
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    but 40^2 - 4(16)(25) = 0

  20. EmmaH
    • 2 years ago
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    ui havent learned taht i think im suppose to just use the big numbers

  21. amistre64
    • 2 years ago
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    then math it out for maths sake, youll get big numbers that finally reduce and simplify

  22. EmmaH
    • 2 years ago
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    i tried but i keep getting teh wrong number when i have 40^2-4(16)(5) i get 0 right? i get 1,600 -1,600 and i cant square thaty

  23. amistre64
    • 2 years ago
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    what does 1600 - 1600 equal?

  24. EmmaH
    • 2 years ago
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    0 -_- i cant square 0

  25. amistre64
    • 2 years ago
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    sure you can, 0*0 = 0

  26. EmmaH
    • 2 years ago
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    oooooh......

  27. EmmaH
    • 2 years ago
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    so -40 pm 0/32?

  28. amistre64
    • 2 years ago
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    \[\frac{ -40\pm \sqrt{40^{2}-4(16)(25)} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{1600-1600} }{ 2(16) }\] \[\frac{ -40\pm \sqrt{0} }{ 2(16) }\] \[\frac{ -40\pm 0 }{ 2(16) }\] \[\frac{ -40}{ 2(16) }\]

  29. amistre64
    • 2 years ago
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    yes /32 is fine as well

  30. EmmaH
    • 2 years ago
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    THANK YOU SO MUCH!!!!!

  31. amistre64
    • 2 years ago
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    youre welcome

  32. EmmaH
    • 2 years ago
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    so its -40/32?

  33. amistre64
    • 2 years ago
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    yes, which a calculator should be able to put into decimals for you

  34. EmmaH
    • 2 years ago
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    -1.25?

  35. amistre64
    • 2 years ago
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    correct

  36. EmmaH
    • 2 years ago
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    thank you!

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