Question 1112951
A) Factorise: 2^2x -7 • 2^x -8 = 0
B) fund the value(s) for x in part a




      SOLUTION

A) To factorize

 2^2x -7 • 2^x -8 = 0

this  equation can be written as
 
(2^x)² - 7•2^x -8 =0 

let p=2^x .( so any where 2^x is present in the equation,  we replace it with p) . 

by doing so, we would have 

p² -7p -8=0 

because this is a now a quadratic equation,  we would have two possible factors that satisfy the equation above . And they're -8 and 1
 ( by inspection ) . You could however use the quadratic formula if you are not sure which factor to use .

Now, 

p² -7p -8=0 

→ p² -8p + p -8 =0 

→ P(p-8) +1(p-8) =0 

( p-8) is common ,so we pull it out 

→ (p-8)(p+1)=0

∴ ( 2^x -8)(2^x + 1)=0 ...( Factorized ) 
 
B)  we seek the possible real values of x 

from   ( 2^x -8)(2^x + 1)=0

for this to hold , it means that 

 ( 2^x -8) =0  or (2^x + 1)=0

→ 2^x =8  

→ 2^x = 2³

 2 on both sides cancels

we have 

x= 3  

for the other case ,

2^x =-1 
take square on both sides

→ 2^(2x) = 1 

→2^(2x) = 2^0 

→ 2x=0 

→ x= 0 
but this  value doesn't satisfy the

 equation,  hence we pick x=3 as our answer 

∴ ANSWER :X=3