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Maths: Inequalities

  1. Jul 15, 2003 #1
    Can someone help me please with inequalities, I have been attempting this question quite a few times but I still can't get the same answer as the text book.

    What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

    Here is my best attempt:

    p(x^2+x) < 2x^2 + 6x +1

    px^2 + 2p - 2x^2 - 6x -1 < 0

    (p-2)x^2 - 6x + 2p -1 < 0

    For real roots b^2 - 4ac >= 0

    so 6^2 - 4 * (p-2) * (2p-1) >= 0

    36 - (8p^2 - 20p + 8) >= 0

    -8p^2 + 20p + 28 >= 0


    root 1: (-20 + (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
    root 1: -20/-16 + (1296^1/2)/16
    root 1: 400/256 + 1296/256
    = 1696/256

    root 2: (-20 - (400 - (4 * -8 * 28)) ^ 1/2) / -16 >= 0
    root 2: -20/-16 - (1296^1/2)/16
    root 2: 400/256 - 1296/256
    = -896/256

    That's as far as I have got...

    the answer in the textbook is p < -1

    Can you please help.... Thanks a lot.

    Gary.
     
  2. jcsd
  3. Jul 15, 2003 #2
    There's a mistake...

    Not "px^2 + 2p - 2x^2 - 6x -1 < 0"...
    but...px^2 + p*x - 2x^2 - 6x -1 < 0...
    (p-2)*x^2+(p-6)*x-1<0
    delta=p^2-12p+36+4p-8...
    delta=p^2-8p+28...
    I) If p-2>0 then there's no solution...it has a "minimus?"...
    II) If p-2<=0 then...
    delta<0
    p^2-8p+28<0
    p^2-8p+16+12<0
    (p-4)^2+12<0...which is impossible...so there's no such p...
    Are you sure the text is correct ?
     
  4. Jul 15, 2003 #3
    i am really sorry, the question should be:

    p(x^2+2) < 2x^2 + 6x +1

    not

    p(x^2+x) < 2x^2 + 6x +1
     
  5. Jul 15, 2003 #4
    p(x^2+2) < 2x^2 + 6x +1

    Then we have...

    (p-2)*x^2-6*x+(2p-1)<0
    Evidently p-2<0...the function has a "maximum"...
    delta=36-4(2p-1)(p-2)...
    delta=36-4(2p^2-5p+2)...
    delta=36-8p^2+20p-8...

    delta<0...no real solutions...
    -8*p^2+20*p+28<0...
    2*p^2-5*p-7>0...
    dp->delta for this ecuation...
    dp=25+56...
    dp=81...
    p1=(5-9)/4=-1...
    p2=(5+9)/4=7/2...
    because a in this ecuation is >0...p is in (-inf,-1)U(7/2,+inf)...
    now...because p-2<0...p<2...results that p<-1...
    Got it ?
     
  6. Jul 16, 2003 #5
    Evidently p-2<0...the function has a "maximum"...

    How do you know this?
     
  7. Jul 16, 2003 #6
    let f(x)=a*x^2+b*x+c...
    f'(x)=2*a*x+b
    f''(x)=2*a
    where f'(x)=0 you'll have an "extremum"...
    f'(x0)=2*a*x0+b=0...x0=-b/(2*a)...
    f''(x0)=2*a...if f''(x0)>0 then in x0 the concavity is upward...
    else it is downward...so...if a>0 the function looks like this \/...and if a<0 then it looks like this /\...got it now ?
     
  8. Jul 16, 2003 #7
    bogdan, sorry this is the first time I have done this, when you said:

    delta<0...no real solutions...
    -8*p^2+20*p+28<0...

    Then shouldn't the last line be:

    -8*p^2+20*p+28>0...

    In the text book it showed the calculations exactly the same way you did, that's what I don't get.

    Would you mind explaining this?
     
  9. Jul 16, 2003 #8

    HallsofIvy

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    Here's how I would do this problem:

    p(x^2+2) < 2x^2 + 6x +1 is the same as

    px^2+ 2p< 2x^2+ 6x+ 1 or (2-p)x^2+ 6x+ (1-2p)> 0 for all x.

    If this is never 0, then it's discriminant must be negative:

    36- 4(2-p)(1-2p)< 0

    (The inequality is <0 so that the discrimant is negative and there are no real roots.)

    -8p^2+ 20p+ 28< 0
    2p^2- 5p- 7> 0

    2p^2- 5p- 7= 0 has roots -1 and 7/2.

    2p^2- 5p- 7> 0 for x< -1 and x> 7/2.

    Putting x= -2 (less than -1) into the original inequality, we find that p(x^2+2) < 2x^2 + 6x +1 for p< -1- the result we want.
    Putting x= 3 (greater than 7/2) into the original inequality, we find that p(x^2+2) > 2x^2 + 6x +1 for p> 7/2- not what we want.

    The original inequality is true for all x as long as p< -1.
     
  10. Jul 16, 2003 #9
    delta=36-8p^2+20p-8...

    delta<0...no real solutions...
    -8*p^2+20*p+28<0...
    if delta<0 then the function doesn't intersect OX...and because it looks like /\...it must be under OX for every x...got it ?
     
  11. Jul 16, 2003 #10
    Question: What is the set values of p for which p(x^2+x) < 2x^2 + 6x +1 for all real values of x?

    Quote: for all real values of x?

    Then b^2 - 4ac > 0 should be true not

    b^2 - 4ac < 0

    thats what I don't understand, and I really appreciate your help!
     
  12. Jul 16, 2003 #11

    HallsofIvy

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    Well, first, you are back to "x^2+ x" which you told us was incorrect. I'm going to assume you meant p(x^2+2) < 2x^2 + 6x +1.

    As far as "b^2- 4ac" is concerned you have it backwards.

    If ax^2+ bx+ c> 0 (or <0) for all x, then ax^2+ bx+ c must have NO real solutions. That means the two roots that any quadratic has must both be complex: the discriminant, b^2- 4ac must be negative.
     
  13. Jul 16, 2003 #12
    if that delta is non-negative then the ecuation would have at least one solution...that means the inequality would be <=...not <...
    ...draw the graphic of the function...and you'll see...try it with both a<0...and a>0...and delta<0...and delta>0...and it will become clear...
     
  14. Jul 17, 2003 #13
    HallsofIvy, ax^2+bx+c > 0 does not mean both roots are not real.

    consider x^2 + 2x + 3 > 0, the roots are not complex.
     
  15. Jul 17, 2003 #14

    HallsofIvy

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    I may not be very bright, but I can at least use the quadratic formula: the roots of x^2 + 2x + 3 = 0 are (-2+/- sqrt(4-12))/2=
    -1 +/- sqrt(2)i and definitely ARE complex.

    Saying that ax^2+ bx+ c> 0 (or ax^2+ bx+ c< 0) for all x means that it is never EQUAL to 0 for any real x. The roots of the equation MUST be complex.
     
  16. Jul 17, 2003 #15
    sorry that wasn't meant to be offensive.

    but consider x^2 + 3x + 2 = 0, the roots are real.
     
  17. Jul 17, 2003 #16

    HallsofIvy

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    Yes, the roots of x^2 + 3x + 2 = (x+2)(x+1)= 0 are x= -1 and x= -2. It follows FROM THAT that x^2+ 3x+ 2< 0 for -2< x< -1 and that
    x^2+ 3x+ 2> 0 for x< -2 or x> -1.

    The point, once again, is that the roots of ax^2+ bx+ c= 0 are (by definition) places where the value is 0. If the ax^2+ bx+ c> 0 or ax^2+ bx+ c< 0 for ALL X, then there is NO place where the value is 0 ans so no real value of x for which the equation is satisfied: the roots MUST be complex and b^2- 4ac must be negative.
     
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