Question 1112951: A) Factorise: 2^2x -7 • 2^x -8 = 0
B) fund the value(s) for x in part a
Answer by Benbuks(2)   (Show Source):
You can put this solution on YOUR website! 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
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