2x^5-x^2+1=0
can you help me ?
slove it in details
thanks

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

           Hi my lil bunny!

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

Move 1 to the left side of the equation by subtracting it from both sides.

Subtract 1 from -2.

Add 3 to both sides of the equation.

To remove the radical on the left side of the equation, square both sides of the equation.

Simplify each side of the equation.

Multiply the exponents in .

Apply the power rule and multiply exponents,

Cancel the common factor of 2.

Simplify.

Raise 3 to the power of 2.

Solve for x

Move all terms not containing x to the right side of the equation.

Add 5  to both sides of the equation.

Add 9 and 5.

Divide each term by 2  and simplify.

Divide each term in 2x = 14 by 2.

Cancel the common factor of 2.

Divide 14 by 2.

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

Hope this helped you.

Could you maybe give brainliest..?

❀*May*❀

root of f(x) =  -0.7419124700395855 to about 16 figures

Step-by-step explanation:

given

f(x) = 2x^5-x^2+1 = 0

The polynomial is prime, so cannot solve by factoring.

Since it is a 5th degree polynomial, it has at least one real root.

Graphing helps locate where roots are, if more than one.

(refer to first graph)

So there is a real root between -1 and 0.

We will use numerical methods to find the root to a good degree of accuracy.  The technique applies to any univariable function which is differentiable and continuous near the roots.  This requirement is true for all polynomials.

However, we must know approximately where the root is, usually found by graphing.

The formula used is a recursive one, which gives a better approximation (x1) from the initial (x0) one , and can be repeated until the required accuracy is reached.

Here, we see that the slope of the function at the root is quite steep, so convergence will be rapid.

The formula is

x1 = x0 - f(x0) / f'(x0), where

x1 = new approximation

x0 = initial (or previous) approximation

f(x0) = value of function when x=x0

f'(x0) = value of derivative of function when x=x0

For the given function

f(x) = 2x^5-x^2+1 = 0

f'(x) = 10x^4-2x = 2x(5x^3-1)

From the graph of f(x), we can take an initial approximation as

x0 = -1

x1 = -1 - (-2)/12 = -5/6

Repeat using x0=-5/6

x1 = -5/6 - ( 2(-5/6)^5 - (-5/6)^2 + 1 ) / (2(-5/6(5(-5/6)^3-1))

= 0.7565596512088784

Repeat again, multiple times

x1 = -0.7423377914518363

x1 = -0.7419128371988212

x1 = -0.7419124700398593

x1 = -0.7419124700395855

x1 = -0.7419124700395855

So we see that the root of f(x) = x1 = -0.7419124700395855 to about 16 figures

Note that the accuracy of the iterations approximately doubles every time.

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation: