Results 1 to 9 of 9

Math Help - Quadratic expression

  1. #1
    Junior Member
    Joined
    Jul 2010
    Posts
    53

    Quadratic expression

    Hey guys,
    I'm stuck on this tough question ......

    Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

    I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    May 2011
    Posts
    7
    The equation will be 0 only at +- 2/3, so between those, but not including them, the equation will be positive.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, Arka!

    I don't follow your reasoning, but you seem to have the right answer.


    \text{Find the values of }a\text{ which make }\,x^2 - ax + 1 - 2a^2\,\text{ always positive for all real }x.

    If the expression is positive, the parabola has no x-intercepts.
    . . That is, its discriminant is negative.

    We have: . D \;=\; (\text{-}a)^2 - 4(1-2a^2) \:<\:0 \quad\Rightarrow\quad 9a^2-4 \:<\:0

    . . . . . . . . a^2 \:<\:\tfrac{4}{9} \quad\Rightarrow\quad |a| \:<\:\tfrac{2}{3}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2010
    Posts
    53
    Can we not arrive at the result without relying on co-ordinate geometry for making assumptions like "If the expression is positive, the parabola has no x-intercepts" ?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by Arka View Post
    Hey guys,
    I'm stuck on this tough question ......

    Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

    I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
    When, in setting f(x)=0, you found the solutions at

    x=\frac{a\pm\sqrt{9a^2-4}}{2}

    then, for "a" real, you have a single solution (graph is tangent to the x-axis) if 9a^2=4

    You have 2 solutions (graph crosses the x-axis) if 9a^2>4

    In both of the above cases, f(x) does not lie ENTIRELY above the x-axis.

    f(x) lies entirely above the x-axis if 9a^2<4

    This is because the graph is U-shaped, due to the coefficient of
    the highest power of x being positive.

    You could also use calculus to examine the minimum value of f(x).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Arka View Post
    Hey guys,
    I'm stuck on this tough question ......

    Find the values of a which make x^2 - ax + 1 - 2(a^2) always positive for all real x.

    I made the expression equal to 0 ..and then since x is real, the discriminant should be greater than or equal to 0 . I manipulated this inequality to get that x lies between -(2/3) and +(2/3), which is actually the answer given , but I am not satisfied since they said that the the expression should be greater than 0 but I set it equal to 0 .
    Just to throw in my 2 cents one way to do this problem is to complete the square.

    x^2-ax+1-2a^2=x^2-ax+\frac{a^4}{4}-\frac{a^2}{4}+1-2a^2=\left(x-\frac{a}{2} \right)^2+\frac{4-9a^2}{4}

    Now for the quadratic to be positive the term after the square must be positive so we get

    \frac{4-9a^2}{4}> 0 \iff (2-3a)(2+3a)> 0 \iff a \in \left(-\frac{2}{3},\frac{2}{3} \right)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jul 2010
    Posts
    53
    Thanks Guys ,
    That really helped !
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2011
    Posts
    2
    Quote Originally Posted by Soroban View Post
    Hello, Arka!

    I don't follow your reasoning, but you seem to have the right answer.



    If the expression is positive, the parabola has no x-intercepts.
    . . That is, its discriminant is negative.

    We have: . D \;=\; (\text{-}a)^2 - 4(1-2a^2) \:<\:0 \quad\Rightarrow\quad 9a^2-4 \:<\:0

    . . . . . . . . a^2 \:<\:\tfrac{4}{9} \quad\Rightarrow\quad |a| \:<\:\tfrac{2}{3}

    I ma little bit confused in the above expression can u clear about it little more
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by gramboy View Post
    I ma little bit confused in the above expression can u clear about it little more
    Be specific. What part confuses you?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic expression
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 31st 2011, 09:12 AM
  2. divisibility of quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 16th 2011, 03:21 AM
  3. Quadratic Expression?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: April 16th 2010, 02:22 AM
  4. [SOLVED] Quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 7th 2010, 10:32 AM
  5. quadratic expression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 6th 2010, 05:29 PM

Search Tags


/mathhelpforum @mathhelpforum