Question

find all square roots! x^5 + 2x^3 - 3x = 0

HELP PLEASE!!!!!!!

Answer 1

You want to factor it.

Answer 2

i tried but i kept doing getting it wrong

Answer 3

\[
x^5 + 2x^3 - 3x = 0 \implies (x-0)(x^4+2x^2+3) = 0
\]This means \(0\) is a root. then we must find the roots of \(x^4+2x^2+3\)

Answer 4

The trick here is to find the roots of \(x^2\) instead of \(x\). \[
(x^2)^2+2(x^2) + 3 = 0
\]Now notice how it looks like a quadratic equation.

Answer 5

You have coefficients \[
a = 1,b=2, c=3
\]Do you know the quadratic formula?

Answer 6

Then find the roots in terms of \(x^2\).

Answer 7

What are you getting from the quadratic formula?

Answer 8

\[x^5+2x^3-3x=0\]\[x(x^4+2x^2-3)=0\]From here we got 2 equations
****FIRST\[x=0\]****SECOND\[x^4+2x^2-3=0\]\[x^4+2x^2-3+4=4\]Added four both sides is just zero. So, you didn't do anything in the equation. This won't affect the final values of x. \[x^4+2x^2+1=4\]\[(x^2+1)^2=4\]\[\sqrt{(x^2+1)^2}=\pm \sqrt{4}\]\[x^2+1=\pm2\]\[x^2=\pm2-1\]\[\sqrt{x^2}=\sqrt{\pm2-1}\]\[x=\sqrt{\pm2-1}\]From here you got two values of x: \[x=\sqrt{2-1}=1\]\[x=\sqrt{-2-1}=\sqrt{-3}=\sqrt{-1(3)}=3i\]I applied an imaginary number to the second one. But if you are asked to get the values of x, just take \[x=0\]\[x=\]Disregard the value of x that has an imaginary number.

Answer 9

x=1

Answer 10

but their are 5 answers

Answer 11

\[\sqrt{x^2}=\pm \sqrt{\pm2-1}\] \[x=\pm \sqrt{\pm2-1}\]This gives us 4 values of x
\[x=\sqrt{2-1}=1\]\[x=\sqrt{-2-1}=\sqrt{-3}=\sqrt{(-1)(3)}=i \sqrt{3}\]\[x=-\sqrt{2-1}=-1\]\[x=-\sqrt{-2-1}=-\sqrt{-3}=-\sqrt{(-1)(3)}=-i \sqrt{3}\]Then you add the\[x=0\]There! You got 5 values of x