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

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

$(y-4)^2 = 11$

$y-4 = - \sqrt{11} , y-4 = \sqrt{11}$

$y= 4 - \sqrt{11} , 4+ \sqrt{11}$

Now we do the back substitution.

$x^2 = 4- \sqrt{11} \\ x = - \sqrt{4 - \sqrt{11}} , \sqrt{4-\sqrt{11}}$ and

$x^2 = 4+ \sqrt{11} \\ x = - \sqrt{4 + \sqrt{11}} , \sqrt{4+\sqrt{11}}$