If $x + 1 = x^2$ and $x > 0$, then $2x^4$ is

Given: $x + 1 = x^2$ and $x > 0$

$x^2 - x - 1 = 0$

When we have an equation in the form $ax^2 + bx + c = 0$, the quadratic formula gives us the exact solutions.

Using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here: $a = 1$, $b = -1$, $c = -1$

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}$

$= \frac{1 \pm \sqrt{1 + 4}}{2}$

$= \frac{1 \pm \sqrt{5}}{2}$

Since $x > 0$, we take the positive root:

$x = \frac{1 + \sqrt{5}}{2}$

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Key insight: Instead of calculating $x^4$ directly (which would be messy), we can use the original equation $x^2 = x + 1$ to simplify our work.

From $x^2 = x + 1$, we can find $x^4$:

$x^4 = (x^2)^2 = (x + 1)^2$

$(x + 1)^2 = x^2 + 2x + 1$

Since we know $x^2 = x + 1$, we can substitute this:

$x^4 = x^2 + 2x + 1 = (x + 1) + 2x + 1 = 3x + 2$

This works because we're using our original equation to keep simplifying until we get everything in terms of $x$ only.

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Now we can find $2x^4$:

$2x^4 = 2(3x + 2) = 6x + 4$

Substituting $x = \frac{1 + \sqrt{5}}{2}$:

$2x^4 = 6 \cdot \frac{1 + \sqrt{5}}{2} + 4 = 3(1 + \sqrt{5}) + 4$

$= 3 + 3\sqrt{5} + 4 = 7 + 3\sqrt{5}$

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$2x^4 = 7 + 3\sqrt{5}$

This substitution method is much faster than calculating $x^4$ directly. Whenever we have a relationship like $x^2 = x + 1$, we should look for opportunities to substitute and simplify rather than computing high powers directly!