Need Math Help

[CheeseMan42]

here is the problem

Solve the equation 0.5x^4 -5 = x^2 -3 three times - once algebraically, graphically and numerically.

I have already gotten it graphically and numerically but can't get it algebraically. i have tried the quadratic, factoring and substitution. i don't know where im making mistakes. the correct answers are x=1.799 and x=-1.799 i would really appreciate if someone could help me. thanks in advance.

[Silverfox]

http://www.mathsnet.net/

 

try there and click AS/A2 2004

 

i DO know how to do these, but it is one am.....cant think...not getting answers right .... argh!

[CheeseMan42]

well thanks for the link but unfortunately i couldn't find anything. if anyone else knows how to do this then please post it. thanks

[daniel]

are those quadratic equations?

[Squarepi]

.5*x^4-5=x^2-3

.5*x^4=x^2+2

x^4=2*x^2+4

x^4-2*x^2-4=0

let u=x^2

u^2-2u-4=0

 

Then just plug into the quadratic formula, solving for u.

u=x^2

so x will equal the +/- square root of u

[red930]

actually to solve correctly you need to factor the equation and set all factors = to zero. because if you dont you wont get the extraineous roots that come with quadratic equations. its a forth degree polynomial so you need to have 4 extraineous roots from it ie (x+5)*(x-2)*(x+9)*(x-8)=0 then set every factor to zero ie

(x+5)=0 and solve for x

(x-2)=0 and solve for x

(x+9)=0 and solve for x

(x-8)=0 and solve for x

dont forget to use "i" or sqrt(-1) when you need to get extraineous roots from polynomials.

[CheeseMan42]

actually there are only 2 roots, i have checked graphically and thanks to squarepi for the help

[red930]

extraineous roots dont show up on graphs thats why they are extraineous..lol

extraineous roots only come from deriving algebraically.

 

an example would be (x-5)=0 an extraineous root would be -5i ^4 because i*i*i*i=-1

and -1 * -5 = 5 so it would be an extraineous root in the polynomial.