1. ## How to factor the quadratic

2^(2X) - 5 x 2^(X) + 6 = 0

2. Originally Posted by stepho_145
2^(2X) - 5 x 2^(X) + 6 = 0
$\displaystyle 2^{2x} - 5*2^x + 6 = 0$

$\displaystyle (2^x)^2 - 5*(2^x) + 6 = 0$

For simplicity, lets substitute 2^x with y

$\displaystyle y^2 - 5y + 6 = 0$

$\displaystyle (y - 3)(y-2)= 0$

Now anti-substitute

$\displaystyle (2^x - 3)(2^x - 2)= 0$

3. Originally Posted by stepho_145
2^(2X) - 5 x 2^(X) + 6 = 0
Let $\displaystyle y = 2^x$.

Then this equation becomes
$\displaystyle y^2 - 5y + 6 = 0$

Can you factor this?

(By the way, I strongly discourage the use of "x" as a multiplication symbol at this point in your education. It looks too much like the variable x. You can use the * for multiplication if you need a symbol.)

-Dan

4. Originally Posted by stepho_145
2^(2X) - 5 x 2^(X) + 6 = 0
put $\displaystyle y=2^x$, then your quadratic becomes:

$\displaystyle y^2 - 5 y + 6 = 0$

which obviously factorises to:

$\displaystyle (y-2)(y-3)=0$

or reversing the substitution:

$\displaystyle (2^x-2)(2^x-3)=0$

RonL