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# Find all real solutions for \$x\$ in \$\$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . \$\$

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Find all solutions for \(x\) in \( 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . \) Prove your answer

May 31, 2019
edited by Guest  Jun 1, 2019

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I'm pretty sure that you mistyped that; the answer is unknown.

Jun 1, 2019
edited by CuriousDude  Jun 1, 2019
#2
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I checked and there are no mistypes.  I have changed it to latex so you can look at it easier.

Guest Jun 1, 2019
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Oh, the \(LaTeX\) version is much clearer.

Simply both sides first so you can isolate the variable.

Remember, isolate, isolate, isolate.

\(x=-1,0,1\)

\(x\) can be -1, 0, or 1. CuriousDude  Jun 2, 2019
#4
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The answer is correct but is it possible that you could go into further detail on how you ended up with your answers?

Guest Jun 2, 2019
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\(2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 \\ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2(2^{x} -1) \\ 2(2^x- 1) x^2- 2(2^{x} -1) =- (2^{x^2}-2)x \\ 2(2^x- 1) (x^2-1) =- (2^{x^2}-2)x \\\)

By inspection,

If both sides equal zero then they both must be the same.

If x=0 both sides are zero so that is a solution.

If x= +1 or -1 they are solutions too.

If x is bigger than 1 then the left side is positive and the right side is negative so no answers there

If x is less than -1 then the left side is negative and the right side is positive so no answer there either.

So could there be any more answers between -1 and +1?

I do not think so but i do not know how to prove it.

Maybe CuriousDude can show us both. Jun 7, 2019