What is the sum of the roots of the equation #4^x - 3(2^(x+3)) + 128 = 0#?

2 Answers
Nov 29, 2016

Given equation

#4^x-3(2^(x+3))+128=0#

#=>(2^2)^x-3(2^x*2^3)+128=0#

#=>(2^x)^2-3(2^x*8)+128=0#

Taking #2^x=y# the equation becomes

#=>y^2-24y+128=0#

#=>y^2-16y-8y+128=0#

#=>y(y-16)-8(y-16)=0#

#=>(y-16)(y-8)=0#

So #y =8 and y =16#

when #y=8=>2^x=2^3=>x=3#

when #y=16=>2^x=2^4=>x=4#

Hence roots are #3 and 4#

So the sum of the roots is #=3+4=7#

Nov 29, 2016

Answer:

#7#

Explanation:

If #p(x)=(x-a)(x-b)=x^2-(a+b)x+ab#

the #x# coefficient is the sum of roots.

In #(2^x)^2-24 cdot 2^x+128# we have that

#24# is the sum of #r_1# and #r_2# such that

#(2^x-r_1)(2^x-r_2)=0#

Also we have #r_1r_2=2^7=2^3 2^4# and
#r_1+r_2=3 cdot 2^3= 2^3+2^4#

then

#r_1=2^3->x_1=3# and
#r_2=2^4->x_2=4# so

#x_1+x_2=7#