What is the sum of the roots of 2^(2x^2-4x-6)=1?

Answer 1

2^(2x^2 -4x -6) = 2^0
=> 2x^2 -4x -6 = 0
=> 2x^2 +2x -6x -6 = 0
=> 2x(x+1) -6(x+1) = 0

x = 3 or -1
3 + (-1) = 2

Answer is 2

Answer 2

Write 2^y=1, where y=2x^2-4x-6.

Then y=0, so we are trying to solve 2x^2-4x-6=0.

This is a quadratic. We can divide by 2 to give:

x^2 - 2x - 3 = 0

So (x-3)(x+1)=0, and thus x=3 or -1.

The sum of the roots is 3+(-1) = 2.

Answer 3

If 2^(2x^2-4x-6)=1, then 2x^2-4x-6 = 0.

Divide by 2: x^2 - 2x -3 = 0
Factorize: (x + 1) (x - 3) = 0

So x = -1 or x = 3, and the sum of the roots is 2

Answer 4

for the quadratic to be a root, it must equal zero. Then we can factor out a two, leaving
x^2 - 2x - 3 =0 which then factors to (x-3)(x+1). The roots are 3 and -1 which sums to two.

Answer 5

2x^2 - 4x - 6 = 0
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = -1, 3
3 - 1 = 2