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anonymous

  • 5 years ago

using quadratic equation how would i solve x^2+13x+22=0 ?

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  1. anonymous
    • 5 years ago
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    Do you know what a,b and c are?

  2. anonymous
    • 5 years ago
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    my guess is that a = x^2 b= 13x and c= 22 ?

  3. anonymous
    • 5 years ago
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    a is the number in from of the x^2 = 1 b is the number in from of the x = 13 c is the constant = 22 Do you know the quadratic formula?

  4. anonymous
    • 5 years ago
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    OOPS "in front"

  5. anonymous
    • 5 years ago
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    yes i do -b plus or minus square root radical b squared minus 4ac all over 2a ?

  6. anonymous
    • 5 years ago
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    Good.. so you put those in (I like to read it the opposite of b), it helps when you put in values.

  7. anonymous
    • 5 years ago
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    -13 +- sqrt(13^2 - 4(1)(22)) all over 2

  8. anonymous
    • 5 years ago
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    Do the b^2 - 4ac next

  9. anonymous
    • 5 years ago
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    so 169-88 ?

  10. anonymous
    • 5 years ago
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    Yes which = 81 -13 +- sqrt(81) all over 2

  11. anonymous
    • 5 years ago
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    Now... is the sqrt (81) an integer "a perfect number"

  12. anonymous
    • 5 years ago
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    so x equals 13 and 9 ?

  13. anonymous
    • 5 years ago
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    No I will help you

  14. anonymous
    • 5 years ago
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    thank you so much

  15. anonymous
    • 5 years ago
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    sqrt(81) = 9 so you now have (-13 +- 9)/2 Your two answers are (-13 + 9)/2 = -4/2 = -2 Other answer (-13 - 9)/2 = -22/2 = -11

  16. anonymous
    • 5 years ago
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    i get it now... omg thank you so much!

  17. anonymous
    • 5 years ago
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    NP

  18. anonymous
    • 5 years ago
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    Be careful when the stuff under the sqrt root sign does not come out "perfectly" then there are more steps to do... If you have another one that is different.. just post and I'll help

  19. anonymous
    • 5 years ago
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    i have like 3 more but im going to try them on my own first and if i get stuck ill post it (: thanks!

  20. anonymous
    • 5 years ago
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    ok so like you said i theres more steps for ones that are not "perfectly" well i got -7+- sqrt -33/ 2

  21. anonymous
    • 5 years ago
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    i know -33 is not a perfect square so what would i do next?

  22. amistre64
    • 5 years ago
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    factor out any "perfect" squares from 33.....there aint none.. so leave it as is

  23. anonymous
    • 5 years ago
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    thanks!

  24. amistre64
    • 5 years ago
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    if you got sqrt(-33) as an answer, then the only solution available is to go to complex numbers; but usually they just want to know if there are any "real" numbers as a solution. sqrt(-33) has no "real" number solutions because any 2 numbers when multiplied together will give you a (+)positive result

  25. anonymous
    • 5 years ago
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    okay thank you (: so i need help with another problem please? It says to solve by factoring or using quadratic equation; the problem given is (x-6) (5x+9) = 0 how would i do this?

  26. amistre64
    • 5 years ago
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    ok... solve by factoring OR quad formula right?

  27. anonymous
    • 5 years ago
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    yes :)

  28. amistre64
    • 5 years ago
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    since the equation is ALREADY in a factored form; we use the rule that anything times zero = 0. that means that when (x-6) = 0 we have a winner; OR when (5x+9) = 0 we have another winner. So we can get 2 solutions to it, can you tell me what x has to equal to get us a zero?

  29. anonymous
    • 5 years ago
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    x= 6 and 9/5ths ?

  30. amistre64
    • 5 years ago
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    your "numerals" are correct, but check the "signs" for me.

  31. anonymous
    • 5 years ago
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    x= 6 x= -9/5ths ?

  32. amistre64
    • 5 years ago
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    exactly, that was very good :)

  33. anonymous
    • 5 years ago
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    thanks im trying...i was gone from school alot because my parents are going through divorce and its been causing me to get very sick so i've been trying to get caught up on make up work

  34. anonymous
    • 5 years ago
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    thanks for your help though!

  35. amistre64
    • 5 years ago
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    youre quite welcome; and I hope you find better days ahead.

  36. anonymous
    • 5 years ago
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    That problem you were doing with -33, can you post the actual problem?

  37. anonymous
    • 5 years ago
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    x2+7x-18=0

  38. anonymous
    • 5 years ago
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    b^2 - 4ac = 7^2 - 4(1)(-18) = 7^2 = 49 -4(1)(-18) = +72 So 49 + 72 = 121 You will find this is a "perfect square root - no decimals" Let me know what you get after this

  39. anonymous
    • 5 years ago
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    a = 1 b = 7 c = -18

  40. anonymous
    • 5 years ago
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    wouldn't it be -72?

  41. anonymous
    • 5 years ago
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    No... think of it as a (-4) times (a)(c) if you do it your way then you have to think of it as 49 - (-72) which then you have to turn into a + sign... much easier if you use a (-4)

  42. anonymous
    • 5 years ago
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    oooo okayy thank youu!

  43. anonymous
    • 5 years ago
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    Let me know what you get for answers

  44. anonymous
    • 5 years ago
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    49-72 = -23 right?

  45. anonymous
    • 5 years ago
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    or do i add 49 and 72 ?

  46. anonymous
    • 5 years ago
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    7^2 - 4(1)(-18) 49 + 72 = 121 Look at the - between the 7^2 and the 4 we are basically going to turn that into a + and put the - with the 4 so 7^2 + (-4)(1)(-18) = 49+72 = 121

  47. anonymous
    • 5 years ago
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    so then its going to be -7+ 11/2 and then -7 -11/2??

  48. anonymous
    • 5 years ago
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    Yes.. do the (-7 + 11) first then /2 and (-7 - 11) then /2

  49. anonymous
    • 5 years ago
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    answers are 2 and -9 ?

  50. anonymous
    • 5 years ago
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    Yes... Do you have another one?

  51. anonymous
    • 5 years ago
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    thank you! yes i have one more a^2-3a = 6 = 0

  52. anonymous
    • 5 years ago
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    OK, what is the a,b, and c Remember.... when it says to solve and you have like an x^2, IT MUST EQUAL ZERO FIRST BEFORE YOU FIND YOUR A,B, and C This one does, I am just making sure you know that.

  53. anonymous
    • 5 years ago
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    so is it going to be -3 +- sqrt 9-24/ 2 ? 9-24 is -15 which isn't perfect so what would i do next ?

  54. anonymous
    • 5 years ago
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    (before I check that... you had an equal sign before the 6, is it a plus or a minus

  55. anonymous
    • 5 years ago
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    plus sorry

  56. anonymous
    • 5 years ago
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    OK, remember a,b and c are the numbers in front of the x^2, the x and the number... KEEP the SIGNS of THOSE NUMBERS with the NUMBERS. a = 1 b = -3 c = 6 Are you in algebra I or II? Have you heard of imaginary numbers yet? i's

  57. anonymous
    • 5 years ago
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    algebra 1 and yes a little bit

  58. anonymous
    • 5 years ago
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    I am surprised that this is your next problem.. If you are sure you entered it correctly. 3 +- sqrt(9-24)/2 3 +- sqrt(-15)/2 When you have a "-" underneath the sqrt sign you get rid of it by bringing out an "i" in front of the sqrt sign. Notice it is 3 at the beginning because b was -3 3 + i sqrt(15) all over 2 3 - i sqrt(15) all over 2

  59. anonymous
    • 5 years ago
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    im lost with the "i" and "-"

  60. anonymous
    • 5 years ago
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    Are you sure your problem was correct? I doubt you have learned about imaginary numbers. that is why I think your problem may be incorrect.

  61. anonymous
    • 5 years ago
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    no the problem is correct its an extra credit problem

  62. anonymous
    • 5 years ago
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    I would leave it as -15 under the square root... 3 + sqrt(-15) /2 3 - sqrt(-15) /2 Hey.. I read about your family situation. I am a teacher in a school.. I want you to know that Jesus knows what you are going thru and he wants to be there for you. He loves you more than anyone you know. If you ever want me to help you with math, or just "talk", you can email me at amcbeth328@gmail.com My name is Andrea McBeth. Keep working hard. I am very pleased to see that you are trying to learn this.. That is commendable!!! I will be praying for you and your situation.

  63. anonymous
    • 5 years ago
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    thank you so much mam! god bless <3 i will be sure to add your email to contacts (:

  64. anonymous
    • 5 years ago
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    You can call me Andrea :-) by the way.. I saw another of your posts and the answer was not completely correct. How do you factor 2x^4 - 32x^2 You factor out the 2x^2(x^2 - 16) But the (x^2 - 16) can be factored into (x + 4)(x - 4) that is the difference between two squares. So the answer is 2x^2(x + 4)(x - 4)

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