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cdelomas

  • one year ago

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.

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  1. sauravshakya
    • one year ago
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    where is the table?

  2. cdelomas
    • one year ago
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    it is in the question

  3. sauravshakya
    • one year ago
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    2x^2-8=0 2x^2=8 x^2=4 x=-2 or 2

  4. sauravshakya
    • one year ago
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    got it?

  5. cdelomas
    • one year ago
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    is that forthe equation a.

  6. sauravshakya
    • one year ago
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    yes

  7. cdelomas
    • one year ago
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    what about b and c

  8. sauravshakya
    • one year ago
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    Did u understand it?

  9. cdelomas
    • one year ago
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    yes for thef irst one I did

  10. sauravshakya
    • one year ago
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    2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 Now, IF x^2<(+-a)^2 then -a<x<a

  11. sauravshakya
    • one year ago
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    SO, can u complete it?

  12. cdelomas
    • one year ago
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    no do not get that one

  13. sauravshakya
    • one year ago
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    2x^2-8<0 2x^2<8 x^2<4 x^2<(+-2)^2 IS it OKAY up to here?

  14. cdelomas
    • one year ago
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    yes

  15. cdelomas
    • one year ago
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    would it be 4

  16. sauravshakya
    • one year ago
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    IF x^2<(+-a)^2 then -a<x<a

  17. sauravshakya
    • one year ago
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    try to use it

  18. cdelomas
    • one year ago
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    I do notg et it

  19. sauravshakya
    • one year ago
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    x^2<(+-2)^2 -2<x<2

  20. sauravshakya
    • one year ago
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    got it?

  21. cdelomas
    • one year ago
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    yes but how do i figure it out

  22. sauravshakya
    • one year ago
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    HINT: IF x^2<(+-a)^2 then -a<x<a

  23. sauravshakya
    • one year ago
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    It is like formula

  24. cdelomas
    • one year ago
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    howdo i figurethef o rmula out

  25. sauravshakya
    • one year ago
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    U want me to derive it?

  26. cdelomas
    • one year ago
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    what does that mean

  27. cdelomas
    • one year ago
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    hello what doyou m ean

  28. sauravshakya
    • one year ago
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    I am sorry but I guess u need to revise your algebra once

  29. cdelomas
    • one year ago
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    soyoucannoth elp meany further

  30. sauravshakya
    • one year ago
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    Sorry not so good at explaining things

  31. cdelomas
    • one year ago
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    ok Iwil l try someelse

  32. sauravshakya
    • one year ago
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    @mathslover can help u?

  33. sauravshakya
    • one year ago
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    He is good at explaining

  34. mathslover
    • one year ago
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    I am well at explanation :)

  35. cdelomas
    • one year ago
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    socanyouhelpme mathsolver

  36. mathslover
    • one year ago
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    I will be right back ... wait for 2 minutes please

  37. wio
    • one year ago
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    \[ 2x^2-8<0 \\ 2x^2<8 \\ x^2<4 \\ \sqrt{x^2}<\sqrt{4} \]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.\[ |x| < 2 \]

  38. wio
    • one year ago
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    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.

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

  40. wio
    • one year ago
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    One thing worthy of committing to memory is that: \[ |x| < a \implies -a < x < a \\ |x| > a \implies x < -a,\quad x > a \]

  41. cdelomas
    • one year ago
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    so for equation b is it squareroot x2>squareroot 4.

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