Question

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

Answer 1

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`

Answer 2

`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`