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anonymous

  • 5 years ago

if both the roots of x^2-(p-4)x+2e^(2log p)-4=0 are negative then what are the possible range of values of p

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  1. anonymous
    • 5 years ago
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    x^2-(p-4)x+2P^2-4=0

  2. anonymous
    • 5 years ago
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    if both are -ve so p-4<0 and p<4 also p>0 as it is in ln.

  3. amistre64
    • 5 years ago
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    i was on the cusp of saying something similar :)

  4. anonymous
    • 5 years ago
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    but agin diccrim. will be +ve.

  5. anonymous
    • 5 years ago
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    why should p-4 be less than zero if the roots are negative?

  6. amistre64
    • 5 years ago
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    (r+m)(r+n) amkes the roots both -

  7. anonymous
    • 5 years ago
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    (p-4)^2>4(2p^2-4) 7p^2+8p-32<0

  8. amistre64
    • 5 years ago
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    +m+n = + result for -(p-4)

  9. amistre64
    • 5 years ago
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    -p+4 >/ 0 -p >/ -4 p </ 4

  10. amistre64
    • 5 years ago
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    2e^(2log p)-4 >/ 0 since +m(+n) = +result

  11. anonymous
    • 5 years ago
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    not following you guys..:-(

  12. anonymous
    • 5 years ago
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    what is your prob?????????

  13. amistre64
    • 5 years ago
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    suppose we got: (r-4)(r-6) are the zeros here + or - ?

  14. anonymous
    • 5 years ago
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    dipankar: u said "if both are -ve so p-4<0" why is this so?

  15. anonymous
    • 5 years ago
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    because if ax^2+bx+c=0 has roots m,n then m+n=-b/a.

  16. anonymous
    • 5 years ago
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    here b=-(p-4), a=1 and two roots are-ve so m+n=p-4 is -ve.

  17. amistre64
    • 5 years ago
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    2e^(2log p)-4 >/ 0 2e^(2log p) >/ 4 e^(2log p) >/ 2 ln(2 log(p)) >/ ln(2) ln(2) + ln(log(p)) >/ ln(2) ln(log(p)) >/ 0

  18. anonymous
    • 5 years ago
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    ok...got it now.. so whats nxt?

  19. amistre64
    • 5 years ago
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    ln(ln(0)/ln(10)) >/ 0 ln(ln(p)) - ln(10) >/ 0 i think this is how to do it :)

  20. anonymous
    • 5 years ago
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    p<4 also p>0 is that the range?

  21. anonymous
    • 5 years ago
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    no you also have to satisfy that D>=0

  22. anonymous
    • 5 years ago
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    i.e. 7p^2+8p-32<0

  23. anonymous
    • 5 years ago
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    if this gives some value of p that is less than 4 but>0 that will be your range.

  24. amistre64
    • 5 years ago
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    ln(ln(0)/ln(10)) >/ 0 ln(p)/ ln(10) >/ e^0 = 1 ln(p) >/ ln(10) p >/ 10 ok, what if anything did i do wrong

  25. amistre64
    • 5 years ago
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    ln(p)/ ln(10) >/ e^0 = 1 log(p) >/ 1 ....same result for me that way.....

  26. anonymous
    • 5 years ago
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    what is your problem now?????????

  27. anonymous
    • 5 years ago
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    a<0, b<0 a+b<0, ab>0 therefore p-4<0 => p<4; 2p^2-4>0 =>p^2>2 therefore p>sqrt{2} or, p<-sqrt(2) so, p\[p \in(\sqrt{2},4) [p>0]\] again both roots are -ve. b^2-4ac>=0 (p-4)^2-4(2p^2-4)>=0 =>7p^2+8p-32<=0 therefore \[\sqrt{2}<p<(4/7)(\sqrt{15}-1)\] did i get it right?

  28. anonymous
    • 5 years ago
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    yes..

  29. anonymous
    • 5 years ago
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    what are the roots of the eqn 7p^2+8p-32<=0

  30. anonymous
    • 5 years ago
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    can you tell me please.

  31. anonymous
    • 5 years ago
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    (4(+-)sqrt(230))/7

  32. anonymous
    • 5 years ago
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    how????????? its wrong.

  33. anonymous
    • 5 years ago
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    (-8(+-)sqrt(960))/7

  34. anonymous
    • 5 years ago
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    yea i factored 2 out

  35. anonymous
    • 5 years ago
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    n yea it should be -ve 4 up there

  36. anonymous
    • 5 years ago
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    hey how did u get ur denominator to be 7?? shouldnt it be 14??

  37. anonymous
    • 5 years ago
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    sorry 14.

  38. anonymous
    • 5 years ago
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    and also it sqrt(920) nt 960..

  39. anonymous
    • 5 years ago
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    why? 32*4*7+64

  40. anonymous
    • 5 years ago
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    yea..ryt..my bad

  41. anonymous
    • 5 years ago
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    wrong!, what is all this noobery

  42. anonymous
    • 5 years ago
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    it says both the roots are negative , not that they are complex

  43. anonymous
    • 5 years ago
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    it says both the roots are negative , not that they are complex

  44. anonymous
    • 5 years ago
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    if both the roots are negative that means that the sum must be negative, and the product must be positive

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