1. ## 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?

2. Originally Posted by classicstrings
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...

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

3. 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?

4. Originally Posted by classicstrings
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

5. One more question - how would you work it out algebraically? Is it possible to use the discriminant to work it out?

6. Originally Posted by classicstrings
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

7. 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

8. Clever! I like it.

-Dan