# How do you figure out how many real number solutions x^4+4x^3-8x^2+2 has?I am learning calculus. I can't figure out how to find out even 1 solution. Is it because there aren't any, or is it...

How do you figure out how many real number solutions x^4+4x^3-8x^2+2 has?

I am learning calculus. I can't figure out how to find out even 1 solution. Is it because there aren't any, or is it something I'm just not understanding? Please explain!

beckden | High School Teacher | (Level 1) Educator

Posted on

First graph the equation:

You can see there are 4 real solutions because the curve intersects the real x axis at 4 points.  Somewhere around -5.5, -0.5, 0.5 and 1.2.   You can use Newton's method to find these roots:

`x_(i+1) = x_i - (f(x_i))/(f'(x_i)) = x_i - (x_i^4+4x_i^3-8x_i^2+2)/(4x_i^3+12x_i^2-16x_i)`

trying `x_0 = -5.5` we do 4 interations to get

`x_1= -5. 4554196` , `x_2= -5. 4543853` , `x_3= -5. 4543847` ,
`x_4= -5. 4543847`

So first root is `~~` -5.4543847

trying `x_0 = -0.5` we do 4 interations to get

`x_1= -0.45833333, x_2= -0.4560571, x_3= -0.45605037`

` x_4= -0.45605037`

So our second root is ` ``~~` -0.45605037

You can work out the other two x=0.62595760, x=1. 2844775, along with x=-5.4543847 and x=-0.45605037 are the solutions to the above equation.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

In this case Rolle's string can be very helpful. To create this string, you need to determine the zeroes of the 1st derivative of function:

f'(x) = 4x^3 + 12x^2 - 16x

Now, we'll cancel it:

4x^3 + 12x^2 - 16x = 0

We'll divide by 4:

x^3 + 3x^2 - 4x = 0

We'll factorize by x:

x(x^2 + 3x - 4) = 0

We'll cancel each factor:

x = 0

x^2 + 3x - 4 = 0

x^2 + 3x - 3 - 1 = 0

(x^2 - 1) + 3(x-1) = 0

(x-1)(x+1) + 3(x-1) = 0

(x-1)(x + 1 + 3) = 0

x - 1 = 0 => x1 =1

x + 4 = 0 => x2 = -4

The roots of the 1st dericative are -4,0 and 1.

Now, we'll create the Rolle's string. The important elements of the string are the values of the function at the x = -4 ; x = 0 and x = 1

f(-4) = (-4)^4 + 4*(-4)^3 - 8*(-4)^2 + 2

f(-4) = 256 - 256 - 128 + 2

f(-4) = -126

f(0) = 2

f(1) = 1 + 4 - 8 + 2

f(1) = -1

We'll arrange the zeroes of derivative on x axis.

`-oo` -4       0    1     +`oo`

-`oo` f(-4)  f(0)  f(1)  + `oo`

-126     2     -1

If there are changes in sign at two consecutive values of the function, that means that between -126 and 2, there is a value of x for f(x) = 0. That means that the graph of function will intersect x axis between -4 and 0.

We notice another change of sign between 0 and 1. There are changes in sign between -`oo` and -4 and between 1 and +`oo` .

Therefore, all the 4 roots of the function are real roots and they belong to the next 4 intervals:

(-`oo` ; -4) ; (-4 ; 0) ; (0 ; 1) ; (1 ; `oo` ).

thinker41 | eNotes Newbie

Posted on

Thank you for your help! I understand it now!