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
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
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