4x^4-13x^2+9 (1 Viewer)

[ta1g]

How do i factorise this? Can someone show me how step by step.

 

[香港!]

4x^4-13x^2+9=0
just treat it as a quadratic equation

so x²=[13+-sqrt(13²-4(4)(9) )]\8
=(13+-5)\8
.: x²=9\4 or x²=1
x=+-3\2 or x=+-1
so
(x-1)(x+1)(x-3\2)(x+3\2)=0
(x²-1)(x²-9\4)=0
x^4-13\4 x²+9\4=0
4x^4-13x²+9=0

 

[withoutaface]

How do i factorise this? Can someone show me how step by step.

Treat x^2 like x:
x^2 = (13 +/-5) / 8

Hence (x^2- ( 9/4))(x^2- 1)
(x-3/2)(x+3/2)(x+1)(x-1)

 

[joey_prince42]

Try doin this instead

I agree with the previous posts about treating the quartic (p(x) of degree 4) as quadratic. But instead of using of solving the roots using the quadratic formulae you are better off splitting the middle term or the cross method depending on what you feel comfortable with - I prefer the splitting it lol.

Let u = x^2

Therefore the expression becomes 4u^2-13u +9

now multiply the constant by 4 since that is the leading coeffecient, which gives you 36. Now find two numbers that when added equal -13 and multiplied give 36.

These two numbers are -4 and -9.

Thefore "splitting the middle term", the expression becomes 4u^2 - 4u -9u + 9

Then factorise:

4u(u-1) -9(u-1)

which becomes:

(4u-9)(u-1)

then re-substituing x^2=u

(4x^2-9)(x^2-1)

which becomes:

(2x-3)(2x+3)(x-1)(x+1)

using the difference of two squares.

I hope this is an easier way...

Regards,

Joey

 

[who_loves_maths]

here's a third way of doing it just for fun.

solve it via inspection... usually you always only need to try the numbers +/- 1... which happens to work in this case. {you should immediately have noticed the coefficients when you saw the equation anyways ... (4 + 9 - 13) = 0 }
then simply use polynomial division

 

[acmilan]

Theres a fourth method to solve quartic equations:

After rigorous proof, it can be shown that cubic and quartic equations have formulae that can solve those respective equations. The links show the methods to do so. Since youre only looking for real roots, you ignore the parts that give the imaginary roots (hence the existence of cosine in that formula but not sine, since sine represents the imaginary roots which can be seen using e^i@, but thats beyond this course)

 

[Emma-Jayde]

then simply use polynomial division

Polynomial division isn't in the 2u course though, is it?
And I don't believe imaginary roots are either...

 

[pLuvia]

just make x^2 = a or another pronumeral, then ull have

4a^2 - 13a + 9 =0

then wehn u factorise you just make the solutions = x^2

then you can find x

 

[who_loves_maths]

^ lol, yea i think we've all pretty much acknowledged that method already

 

[acmilan]

Polynomial division isn't in the 2u course though, is it?
And I don't believe imaginary roots are either...

Nothing wrong with a little extra knowledge, I myself was a 2uniter last year

 

[who_loves_maths]

here's a fifth method that's even more general (but equivalent) to acmilan's:

use the general formula for all quartic polynomials!

lol... but maybe that's stretching 2u maths a tad too much ...

 

[Emma-Jayde]

Nothing wrong with a little extra knowledge, I myself was a 2uniter last year

I'm not saying there is!
But don't you think it would be better not to confuse people who will see 'polynomial division' or 'imaginary roots' and wonder why they haven't leanrt them?
They might start to panic!

 

[who_loves_maths]

^ are you one of them?

sorry for alarming you if you are...

 

[Emma-Jayde]

haha, I'm definately not!
But I know a lot of people (2 uniters anyway) that would be quite worried by the prospect...Especially when you talk about them in 'mathematics', and not ext 1 or 2!