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Mixed exponential power (Posted on 2025-05-06)

Solve following equation

(x²-x-1)e^x²+2x-3+(x²+2x-3)e^x²-x-1 = 2x²+x-4

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

Comment 1 of 1

To attempt a general analytic solution to this seems impossible unless there is a trick or at least a short cut.

Keeping in mind that e^0=1 and 0^k is always 0 (unless k is zero).

Lets check what the roots of the quadratics are

(x^2-x-1) (1 ± √5)/2 roots are {φ, 1-φ}

(x^2+2x-3) (x+3)(x-1) roots are {1, -3}

So I'm guessing probable solutions: {1, -3, φ, 1-φ}

Plugging in the numbers works for each, so these are the four solutions.

Posted by Larry on 2025-05-06 14:26:15

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