Really quick problem!

J

Joyce

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Feb 16, 2006
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I know this is my second post but it's my last I promise! (studying for a test).


a)2^2x-9(2^x)+8=0

b) 4^2x-17(4^x)+16=0


I know ur supposed to factor but not sure how.........
 
Joyce said:
I know this is my second post but it's my last I promise! (studying for a test).
a)2^2x-9(2^x)+8=0
b) 4^2x-17(4^x)+16=0
I know ur supposed to factor but not sure how.........
Click to expand...

ur = you are ? or u times r ? :shock:

Your equations look kind of suspicious;
like, is 2^2x = (2^2)x which equals 4x, or 2^(2x) ?

Check them please.
 
Joyce said:
a)2^2x-9(2^x)+8=0
b) 4^2x-17(4^x)+16=0
Click to expand...
Is the first equation meant to be as follows?

. . . . .2<sup>2x</sup> - 9(2<sup>x</sup>) + 8 = 0

If so, notice that (2<sup>x</sup>)<sup>2</sup> = 2<sup>2x</sup>, so the equation is a quadratic in 2<sup>x</sup>. Factor (or apply the Quadratic Formula) to solve for the value(s) of 2<sup>x</sup>, and then solve for the value(s) of x.

If you get stuck, or if the above is mistaken, please reply showing your steps.

Thank you.

Eliz.
 
k, i'm stuck.........


2^2x-9(2^x)+8=0

thererfore it's a quadratic equation:

a= 2

b= -18

c= 8

Solve by factoring;

I tried using the quadratic formula but i didn't get whole numbers. The answer at the back of the book was X=3, 0
 
Joyce said:
a= 2, b= -18, c= 8
Click to expand...
You have been given a quadratic in 2<sup>x</sup>. How are you getting these values (above) for the coefficients of 2<sup>2x</sup> and 2<sup>x</sup>?

Please reply showing your steps. Thank you.

Eliz.
 
what is <sub>?


2^2x-9(2^x)+8=0

Factor:

(2^x+1)(2^x+8)

2^x+1=0

2^x= -1


2^x+8=0

2^x= -8

x= 1/8

but tye answers at the back of the book is X=3, 0
 
Joyce said:
what is <sub>?
Click to expand...
An HTML tag.

Joyce said:
2^2x-9(2^x)+8=0
Factor: (2^x+1)(2^x+8)
Click to expand...
Check your work: (y + 1)(y + 8) does not multiply to y<sup>2</sup> - 9y + 8.

You're very close, though.

Eliz.
 
when you factor something I now your supposed to break the equation into two, like:

(x+1)(x-1)

but how do you know which numbers to use?

the answer says X=3,0 but I never get that!
 
Joyce said:
2^2x-9(2^x)+8=0
Factor:
(2^x+1)(2^x+8)
Click to expand...
Should be (2^x - 1)(2^x - 8) = 0
SO: 2^x = 1 or 2^x = 8
x = log(1) / log(2) or x = log(8) / log(2)
x = 0 or 3
 
Maryanne said:
[W]hen you factor something I [know] [you're] supposed to break the equation into two...but how do you know which numbers to use?
Click to expand...
So your class hasn't covered how to factor quadratics yet...?

Eliz.
 
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