Your best bet is to do this numerically:

First: notice that the right hand side of the equation, `e^(x^4)` , is always positive. But the left hand side of the equation, 4x, is negative if x<0.

So, we only need to look for positive roots

I don't know...

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Your best bet is to do this numerically:

First: notice that the right hand side of the equation, `e^(x^4)` , is always positive. But the left hand side of the equation, 4x, is negative if x<0.

So, we only need to look for positive roots

I don't know if you have taken a calculus class, but if you have, you can use the derivative of

`e^(x^4)-4x` to figure out that there can't be any roots larger than 2.

At x=2, we have: `e^(2^4)-4*2>0`

The derivative is

`e^(x^4) *4x^3-4 = 4(e^(x^4) x^3 -1)`

If x>2, this is positive.

So, at x=2 the function is positive and continues to increase, so it is never 0.

Thus, any possible roots are between 0 and 2.

Graphing the left hand side and the right hand side, we can get a better idea of where the solutions are:

They look like they occur at roughly x=.3 and x=1.1

Software gives the solutions as:

`x = .250994 ...` , `x=1.10394 ...`

If you don't want to use software, you have a couple of algorithms.

A good (quicker) on is Newton's method, if you know what that is.

But the easiest algorithm is probably the following:

We will try to find the root that is near .3

We know that 0 is too small to be the root, and 1 is too big (based on the graph)

`e^(0^4)-4*0=1>0`

`e^(1^4)-4*1=-3<0`

So we try half way between 0 and 1, which is .5, and try plugging that in:

`e^(.5^4)-4*.5=-.9355... <0`

This is negative, so now we know the root is between 0 and .5

We try half way between 0 and .5, which is .25, and try plugging that in:

`e^(.25^4)-4*.25=.0039... >0`

This is positive, so now we know that the root is between .25 and .5

We try plugging in .375

etc, etc

Keep doing this proccedure until you have as much precision (as many decimal places) as you need.

Repeat the same procedure for the other root