Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Is it

2 is the the only real solution.

That looks okay now.

That factorization does not look correct.

Try to get all those terms to look like 2^x, that should work.

Is that 1 equation?

That is correct.

10 is not an answer but 13 is.

Solve the first quadratic and you will see why.

Yes, those are correct. You could have derived the exact answers though.

Hi;

Sorry, I am close to a solution of my own and prefer not to look until it fizzles out.

You mean two roots from the quadratic equation? Yes, there will be two answers for y.

The y's in post #352 are changed to 2^x and solved for.

Those are correct for the plugging in.

That is correct! You will get,

You will get the polynomial when you use the substitution of y = 2^x of y^2 - 2y - 1 = 0.

I think the same method to factor but use the quadratic formula to get the roots will get the answer.

I also think the book meant to solve this problem instead.

2^x + 2^-x = 2

Does 0 equal 2?

Please plug in x = 0 and tell me what you get?

Have you copied the problem correctly?

2^x - 2^-x = 2 does not have zero for a root.

That can not be factored, you are correct. We need another way.

Multiply both sides by 2^x and do what you did before.