x^4+100x^2+2304=0?

Question

4 answers

Answer 1

let y = x^2 and solve the quadratic
y^2 + 100 y + 2304 = 0
that will give you 2 answers for y, then just take the +/- square roots of those 2 solutions to get your 4.
You should be able to handle filling in those steps. Good luck!

Answer 2

OK - here you go

Let x^2 = y

Then you have y^2 +100y +2304 =0

Factoring this :

(y +36) (y+64) so the answers are y = -36 and -64.

But we replaced x^2 with y ... so x^2=-36 and x^2=-64

so x = -6i and -8i and 6i and 8i

Checking your answer

(-6i)^4+ 100(-6i)^2 +2304 =0 ?

1296 + -3600 + 2304 = 0 OK and

(-8i)^4 + 100(-8i)^2 +2304 = 0?

4096 + -6400 +2304 = 0 OK!

Works for 6i and 8i too.

Note that when an imaginary number is placed to the 2 power it is -, but when you place it to the fourth power it is a positive since a (-)^2 is a positive.

Hope this helps.

Answer 3

you should get a complex number if this is right

x^4+100x^2+2304=0
(x^2+64)(x^2+36)=0
=> x^2 = -64 or -36

if the equation is x^4 - 100x^2+2304=0
then you should get x^2 = 64 or 36
x= -6, -8 +6 or +8

Answer 4

X^4+100x^2+2304=0
x^4 + 100x^2 + 2500 = 196
(x^2 +50)^2 = (14)^2
x^2 + 50 = +/-14
x^2 = 14 - 50 = -36
x = -6i
x = -8i
x = 6i
x = 8i

Answer 5

let y=x^2
y^2=x^4
y^2+100y+2304=0
solve by completing the square
y^2+100y+2500=-2304+2500
(y+50)^2=196
y=+/- sqrt(196)-50
y1=sqrt(196)-50=-36
y2=-sqrt(196)-50=-14-50=-64
solve for x
x^2=-36
x=+/- 6i
x^2=-64
x=+/- 8i
therefore the roots are 6i,-6i,8i and -8i