Solve by a numerical method, eg the Newton-Raphson iterative method.

This improves an estimate `x_n` to `x_(n+1)` by the formula

`x_(n+1) = x_n - f(x_n)/(f'(x_n))`

where `f(x)` is the function to be solved (equated to zero) and `f'(x)` is its derivative with respect to `x`.

Here `f(x) = x^2e^x - 1`

so that `f'(x) = x^2e^x + 2xe^x - 1 = xe^x(x+2) - 1`

and `f(x)/(f'(x)) = (x^2e^x-1)/(xe^x(x+2)-1)`

First choose a starting point/estimate `x_0` of the solution. Let's try `x_0=1` . Using the iterative formula given above

`x_1 = 0.760`, `x_2 = 0.693`, `x_3 = 0.708`, `x_4 = 0.702` ,

`x_5 = 0.704`, `x_6 = 0.703`, `x_7 = 0.704`, `x_8 = 0.703`,

`x_9 = 0.703`

` `When the answers repeat we stop.

Therefore ` ``x = 0.703` to 3sf

Check there isn't another solution by looking at the graph

The graph tends to infinity for larger x and eventually tends to -1 for smaller x after a maximum of -0.5 at x approx -2.

Using the iterative method of Newton-Raphson x = 0.703 to 3sf

We have

`x^2e^x=1`

Let us assume `x>0`

`log(x^2e^x)=log(1)`

`log(x^2)+log(e^x)=0`

`2log(x)+x=0` (i)

Let consider a function f(x)=2 log(x)+x .

**Solving (i) , and find the zero of f(x) stands same.**

Interpret graph

x=.8 ,red line above x axis

x=.69 green line below x axis

Here approximately we can say x=.705 ( **one can estimate solution by Newton Raphson method).**