A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Last problem! Help for medal. I need help with just one more inequality equation... x^2 - x - 15 > 15

  • This Question is Closed
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x<-5 or x>6

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Would you like me to explain?

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes please

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sure

  5. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @_greatmath7 thanks for providing an explanation in addition to the direct answer. :)

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let's solve your inequality step-by-step. x^2-x-15>15 Let's find the critical points of the inequality. (Subtract 15 from both sides): x^2-x-15-15=15-15 You then receive: x^2-x-30=0 (Factor left side of equation):(x+5)(x-6)=0 (Set factors equal to 0):x+5=0 and x-=0

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thnx @UsukiDoll

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x-6=0

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh I see. Thank you very much! :)

  10. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    o_0 \[\LARGE x^2 - x - 15 > 15\] first we subtract 15 from both sides \[\LARGE x^2 - x - 15-15 > 15-15\] \[\LARGE x^2 - x - 30 > 0\] then we have to factor. our last term is -30 and the middle term is -1 6 x -5 is the only combination that can work for this quadratic equation because 6 x -5=-30 6-5 = 1 Therefore \[\LARGE (x-6)(x+5) > 0\] now we solve \[\LARGE (x-6)(x+5) > 0\] by splitting this up into two cases \[\LARGE (x-6) > 0\] \[\LARGE (x+5) > 0\] solving for an inequality is the same as solving for an equation \[\LARGE x-6 > 0\] \[\LARGE x-6+6 > 0+6\] \[\LARGE x > 6\] \[\LARGE x+5 > 0\] \[\LARGE x+5-5 > 0-5\] \[\LARGE x > -5\] time to switch the inequality sign \[\LARGE x < -5\]

  11. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    on the number line since we have < we have an open circle |dw:1439510309024:dw|

  12. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    for example x < -5 the mouth of < is going on the right, but actually we have to shade the left side on the number line according to the tail similarly x > 6 the > is going left, but actually we have to shade to the right on the number line according to the tail.

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm sorry, but I've gotten a bit lost on the sign switching step. Could you explain that again?

  14. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I got a bit confused on that one too... I think it has something to do with the negative number. I know that if you divide both sides by a negative number the inequality signs switch, so I'm assuming that since we have a negative number as one of the solutions, the inequality signs switch. But what I've read is that only swapping sides, and multiplication/division by negative numbers switch the inequality signs so x > -5 swapping sides -5 < x but it's a standard to have that variable x on the left hand side.

  15. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ah..maybe switching x> -5 by swapping sides might be it? -5 < x but nobody writes it like that even though it's still the correct answer.

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Oops duh. Thank you for your help as well. :)

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    So it would be 6 < x > -5

  18. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x.x I would just leave it as x< -5 and x >6

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What confuses me is how x > -5 turns into x< -5

  20. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it probably because x > -5 doesn't make sense. x can't be greater than -5 the number lines either look like |dw:1439512229503:dw| or |dw:1439512254356:dw|

  21. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ^ that's just an example by the way the first number line's notation is -2 < x < 3 in order from smallest to greatest the second number line's notation is x < -2 or x >3

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If x is greater than 6, wouldn't it also be greater than -5? (I'm sorry if I'm becoming cumbersome. I am just having a hard time understanding)

  23. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no. x > 6 x < -5 http://www.wolframalpha.com/input/?i=solve+x%5E2+-+x+-+15+ >+15 I already mentioned that when putting the inequality on the number line it should either be in the format of -2<x<3 |dw:1439514392007:dw| or x < -2 or x>3 |dw:1439514398312:dw|

  24. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you know what's off? only multiplying or dividing with negative numbers switches the signs.. a part of me wants x> -5 to be that answer. but stupid wolfram has it at x <-5 if I switch the sides on x >-5 and change the inequality sign -5<x which is correct, but the standard is x on the left hand side even though -5 < x is correct.

  25. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    another example Example: 12 < x + 5 If we subtract 5 from both sides, we get: 12 - 5 < x + 5 - 5 7 < x That is a solution! But it is normal to put "x" on the left hand side ... ... so let us flip sides (and the inequality sign!): x > 7 Do you see how the inequality sign still "points at" the smaller value (7) ? And that is our solution: x > 7 source: http://www.mathsisfun.com/algebra/inequality-solving.html

  26. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.