Question

Given:f(x)=x^4-8x^2+5 find all roots to the nearest 0.001

Answer 1

`f(x) = x^4-8x^2+5`

We let x square =y

so x to the power 4 = y square

and we get

`f(y)= y^2-8y+5`

To find the zroes, we need to set f(y) to 0 and solve for y .

`y^2 -8y+5=0`

Now we use completing the square method

`y^2 -8y +16=-5+16`

`(y-4)^2 = 11`

`y-4 = - √(11) , y-4 = √(11)`

`y= 4 - √(11) , 4+ √(11)`

Now we do the back substitution.

`x^2 = 4- √(11) \\ x = - \sqrt{4 - √(11)} , \sqrt{4-√(11)}` and

`x^2 = 4+ √(11) \\ x = - \sqrt{4 + √(11)} , \sqrt{4+√(11)}`