Determine how many, what type, and find the roots for f(x) = x4 + 21x2 − 100.

Question

Rylee88 · Answer

x4 Find the variable (x) _4 + 21 = ____ * 2 = ____ - 100 = ___

kekeman · Answer

Ok i was thinking that the roots were -x+2=0, x=-2, x-2=0, x=2, x-5i=0, x=5i, x+5i=0, x=-5

surjithayer · Answer

it has four roots (being of 4th degree).
put $$x^2=t,x^4=t^2$$
f(x)=0 , gives
$$t^2+21t-100=0$$
$$t^2+25t-4t-100=0$$
$$t(t+25)-4(t+25)=0$$
$$(t+25)(t-4)=0$$
$$(x^2+25)(x^2-4)=0$$
$$(x^2-(-25))(x+2)(x-2)=0$$
$$(x^2-25 \iota^2)(x+2)(x-2)=0$$
$$(x+5 \iota)(x-5 \iota)(x+2)(x-2)=0$$
$$x=5 \iota,-5 \iota,2,-2$$

kekeman · Answer

@rylee88 wrote:

x4 Find the variable (x) _4 + 21 = ____ * 2 = ____ - 100 = ___

Ok i was thinking that the roots were -x+2=0, x=-2, x-2=0, x=2, x-5i=0, x=5i, x+5i=0, x=-5

kekeman · Answer

@surjithayer wrote:

it has four roots (being of 4th degree). put $$x^2=t,x^4=t^2$$ f(x)=0 , gives $$t^2+21t-100=0$$ $$t^2+25t-4t-100=0$$ $$t(t+25)-4(t+25)=0$$ $$(t+25)(t-4)=0$$ $$(x^2+25)(x^2-4)=0$$ $$(x^2-(-25))(x+2)(x-2)=0$$ $$(x^2-25 \iota^2)(x+2)(x-2)=0$$ $$(x+5 \iota)(x-5 \iota)(x+2)(x-2)=0$$ $$x=5 \iota,-5 \iota,2,-2$$

Thank you!

kekeman · Answer

That was super helpful

surjithayer · Answer

yw