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Math Help - Using Discriminant

  1. #1
    Member classicstrings's Avatar
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    Using Discriminant

    For what value/s of b does 2^2x + 2^x + b = 0 have only one real solution?

    Let 2^x = y

    y^2 + y + b = 0

    Discriminant = 1 - 4b, when that is 0, then one real solution;

    b = 1/4

    However this is not the correct answer, why?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    For what value/s of b does 2^2x + 2^x + b = 0 have only one real solution?

    Let 2^x = y

    y^2 + y + b = 0

    Discriminant = 1 - 4b, when that is 0, then one real solution;

    b = 1/4

    However this is not the correct answer, why?
    What would the value of x be for this root?
    2^(2x) + (2^x) + (1/4) = 0

    2^x = [-1 (+/-) sqrt{1 - 4*1*(1/4)}]/2

    2^x = -1/2

    but 2^x can never be negative...

    Your method is okay, but your y is restricted.

    A simpler way would be to simply graph the thing for several values of b. It is then easy to recognize that the solution is b < 0.

    -Dan
    Attached Thumbnails Attached Thumbnails Using Discriminant-exponential.jpg  
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  3. #3
    Member classicstrings's Avatar
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    Thanks for pointing out it is actually quite easy to do by subbing in values for b, I didn't realise that.

    Is there a easy way to do this if it was calculator free?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    Thanks for pointing out it is actually quite easy to do by subbing in values for b, I didn't realise that.

    Is there a easy way to do this if it was calculator free?
    I realized the solution simply by noting that 2^(2x) + 2^x is never negative and for x -> -infinity approaches 0. Since 2^(2x) + 2^x increases monotonically the solution had to be b < 0.

    However unless you are a total Math geek like me you probably wouldn't have thought to analyze it in this manner.

    -Dan
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  5. #5
    Member classicstrings's Avatar
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    One more question - how would you work it out algebraically? Is it possible to use the discriminant to work it out?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by classicstrings View Post
    One more question - how would you work it out algebraically? Is it possible to use the discriminant to work it out?
    Algebraically? Not that I know of. On reflection the best plan of attack I know of is the semi-Calculus approach that I sketched out previously.

    -Dan
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  7. #7
    Senior Member ecMathGeek's Avatar
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    I think it may be possible to do it algebraically. I'll give it a shot:

    Before I begin, you have to know what logarithms are. If you don't then what I'm about to do might not make sense to you.

    For what values of b does 2^2x + 2^x + b = 0 have only one real solution?

    setting 2^x = y is fine, but you must remember to back substitute because 2^x has its own range which is different from a quadratic function's:

    y = (-b+-sqrt(b^2-4ac))/(2a) ... I know you know this.
    y = (-1+-sqrt(1-4b))/2 ... now lets back substitute y = 2^x
    2^x = (-1+-sqrt(1-4b))/2 ... get x by itself by taking log[2] to both sides ... Note: log[2] is my way of saying 'log base 2'
    x = log[2]((-1+-sqrt(1-4b))/2)

    If you know anything about logs you know they have a limited domain: (0,infinity). What this means is the function on the inside of the log[2] cannot be <= 0. Any value other than a positive number will not work. Therefore,
    -1+-sqrt(1-4b) must be > 0 ... let's solve this:

    +-sqrt(1-4b) > 1 ... if it's greater than 1, then it's not negative.
    sqrt(1-4b) > 1 ... square both sides
    1-4b > 1
    -4b > 0 ... divide by negative 4, which flips the inequality
    b < 0
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  8. #8
    Forum Admin topsquark's Avatar
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    Clever! I like it.

    -Dan
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