Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

cdelomas

a table for y=2x^2-8 is given solve each equation. a. 2x^2-8=0,b. 2x^2-8<0,c. 2x^2-8>0.

  • one year ago
  • one year ago

  • This Question is Closed
  1. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    where is the table?

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

    it is in the question

    • one year ago
  3. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    2x^2-8=0 2x^2=8 x^2=4 x=-2 or 2

    • one year ago
  4. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    got it?

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

    is that forthe equation a.

    • one year ago
  6. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    yes

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

    what about b and c

    • one year ago
  8. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    Did u understand it?

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

    yes for thef irst one I did

    • one year ago
  10. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 Now, IF x^2<(+-a)^2 then -a<x<a

    • one year ago
  11. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    SO, can u complete it?

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

    no do not get that one

    • one year ago
  13. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 IS it OKAY up to here?

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

    yes

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

    would it be 4

    • one year ago
  16. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    IF x^2<(+-a)^2 then -a<x<a

    • one year ago
  17. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    try to use it

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

    I do notg et it

    • one year ago
  19. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    x^2<(+-2)^2 -2<x<2

    • one year ago
  20. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    got it?

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

    yes but how do i figure it out

    • one year ago
  22. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    HINT: IF x^2<(+-a)^2 then -a<x<a

    • one year ago
  23. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    It is like formula

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

    howdo i figurethef o rmula out

    • one year ago
  25. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    U want me to derive it?

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

    what does that mean

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

    hello what doyou m ean

    • one year ago
  28. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    I am sorry but I guess u need to revise your algebra once

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

    soyoucannoth elp meany further

    • one year ago
  30. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    Sorry not so good at explaining things

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

    ok Iwil l try someelse

    • one year ago
  32. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    @mathslover can help u?

    • one year ago
  33. sauravshakya
    Best Response
    You've already chosen the best response.
    Medals 1

    He is good at explaining

    • one year ago
  34. mathslover
    Best Response
    You've already chosen the best response.
    Medals 1

    I am well at explanation :)

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

    socanyouhelpme mathsolver

    • one year ago
  36. mathslover
    Best Response
    You've already chosen the best response.
    Medals 1

    I will be right back ... wait for 2 minutes please

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

    \[ 2x^2-8<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4} \]Now it is important to remember that \(\sqrt{\quad}\) mean the "positive square root" (not the negative one) and that \(\sqrt{a^2}\) is the definition of of the absolute value.\[ |x| < 2 \]

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

    To solve absolute value equations, we have to split it up into two cases. Case 1: Assume \(x\) (the thing in the absolute value) is positive. \[ |x| < 2 \\ x<2 \] That was easy. Case 2: Assume \(x\) is negative: \[ |x| < 2 \\ -x < 2 \\ x > -2 \]Remember that if \(x\) is negative, then \(|x| =-x\) because the absolute value bars had to flip the sign to make \(x\) positive. Also remember that when you multiply/divide both sides of an inequality by a negative number, the equality flips. Hence \(<\) became \(>\) in this instance.

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

    Notation allows us to conveniently combine the expressions \(x < 2\) and \(x > -2\) into \(-2 < x < 2\). Does this help, @cdelomas ?

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

    One thing worthy of committing to memory is that: \[ |x| < a \implies -a < x < a \\ |x| > a \implies x < -a,\quad x > a \]

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

    so for equation b is it squareroot x2>squareroot 4.

    • one year ago
    • Attachments:

See more questions >>>

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.