Question

Solve for x.

x^5+8x^3-9x=0

Answer 1

Factor out the lowest common factor within all three terms first. Between \(x^5 ~, ~ x^3~,~ x^1\) which one is the smallest?

Answer 2

x^1

Answer 3

Alright, so factor out an \(x\) from the equation.

Answer 4

x(x^4+8x^2-9)

Answer 5

what happened to the 8x^2?

Answer 6

You are right, \(x^4 +8x^2 -9\).

Answer 7

Solve it like you would any regular quadratic. Treat your \(x^4\) like an \(x^2\)

Answer 8

So find two numbers that multiply to give -9, add to give +8. What are they?

Answer 9

-1, 9

Answer 10

im not too sure what you said about x^4+8x^2-9, how do i make this a regular quadratic?

Answer 11

let \(y=x^2\)

now we have \(y^2+8y-9=0\)

Answer 12

can you solve this for y?

Answer 13

9 -1

Answer 14

Good, so if we use @zzr0ck3r 's method, letting \(y=x^2\), then we ca turn our quadfratic into it's factored form. \(y^2+8y-9=0 \iff (y+9)(y-1)=0\)

Answer 15

quadratic*

Answer 16

Now all we have to do is resubstitute \(x^2\) in for \(y\), and that'll leave us with: \[(x^2+9)(x^2-1)=0\]

Answer 17

is that -3 and i?

Answer 18

\(\pm 3i\)

Answer 19

\[x(x^2+9)(x^2-1)=0\]\[x=0\]\[x^2=-9\implies x=\pm 3i\]\[x^2=1\implies x=\pm 1\]

Answer 20

More specifically, \(x^2 = \sqrt{-9} = \sqrt{-1} \cdot \sqrt{9} = i\cdot(\pm 3) = \pm 3i \)

Answer 21

always remember: \(i= \sqrt{-1}~,~ i^2 = -1\)

Answer 22

Very nicely explained. Thank you!

Answer 23

No problem :)