Question

solve 2x^4-5x^2-12=0

Answer 1

Solving this just like a quadratic, and I will show you a trick.\[2x^4 -5x^2-12=0\]Multiply leading coefficient to end constant. \[x^4 -5x-24=0\]What are two numbers that multiply to give you -24 and add to give you -5?

Answer 2

8 and -3

Answer 3

put x^2=z
Then we have 2z^2-5z-12=0
implies z=4,-3/1
therefore x^2=4,-3/2
from here find x.@domingo

Answer 4

@domingo

Answer 5

what happened to the 5x^2 why did its 2 go away?

Answer 6

And no, 8 and -3 give you -24, +5

Answer 7

So that is why you make the bigger number negative.

-8, 3

Answer 8

I'm lost sorry/

Answer 9

Now we put it in factored form. \[(x^2-8)(x^2+3)=0\]Since it is in factored form, we want to keep to the original equation, thus we must divide out factors by the leading coefficient of our original equation. \[\left(x^2 -\frac{8}{2}\right)\left(x^2+\frac{3}{2}\right)\]

Answer 10

@domingo follow the substitution i mentioned.

Answer 11

so it would be (x^3-8)(x+3)

Answer 12

put x^2=z
Then we have 2z^2-5z-12=0
implies z=4,-3/2
therefore x^2=4,-3/2
from here find x

Answer 13

Once you divide out your fractions, you should get \[(x^2-4)\left(x^2+\frac{3}{2}\right)=0\]Now we simplify multiply the denominator of \(2\) to the variable, and we will have our factored form. \[(x^2 -4)(2x^2+3)=0\]

Answer 14

Do you understand what I did, @domingo ?

Answer 15

up until you factored in the 2, I think your way may be too advanced for me.

Answer 16

So up to this step? \[\left(x^2 -\frac{8}{2}\right)\left(x^2+\frac{3}{2}\right)=0\]

Answer 17

yes thats the one

Answer 18

Now reduce whatever fractoin you can and leave the ones you cannot alone.

Answer 19

fraction*

Answer 20

That would only be the \(-\frac{8}{2}\) correct? Which reduces to \(-4\)

Answer 21

oh ok but I still have to make it equal to zero so (2x^2+3/2)=0

Answer 22

More like \(\dfrac{2x^2+3}{2} =0\) Which will simply become \(2x^2+3=0\)

Answer 23

And all I did there was get rid of the denominator by multiplying it to 0

Answer 24

what would I do without the shortcut?

Answer 25

Without the shortcut, (which might still be considered the shortcut) I would follow @Princer_Jones 's method to reduce the quartic function (4th power function) to a quadratic function (2nd power function)

Answer 26

I am not quite sure what other method would be easiest as I usually use the method I have demonstrated to you.

Answer 27

There are methods like Ferrari's method and also Newton gave a method. Those methods are used when you cannot have quadratic substitution.
Like equations of the form ax^4+bx^3+cx^2+dx+e=0
or ax^4+cx+d=0 etc

Answer 28

princer_jones, i studied yours and it makes a lot of sense thank you

Answer 29

thanks to the both of you