CheeseMan42 Posted November 9, 2004 Posted November 9, 2004 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. Share this post Link to post Share on other sites More sharing options...
Silverfox Posted November 9, 2004 Posted November 9, 2004 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! Share this post Link to post Share on other sites More sharing options...
CheeseMan42 Posted November 9, 2004 Posted November 9, 2004 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 Share this post Link to post Share on other sites More sharing options...
daniel Posted November 9, 2004 Posted November 9, 2004 are those quadratic equations? Share this post Link to post Share on other sites More sharing options...
Squarepi Posted November 9, 2004 Posted November 9, 2004 .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 Share this post Link to post Share on other sites More sharing options...
red930 Posted November 9, 2004 Posted November 9, 2004 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. Share this post Link to post Share on other sites More sharing options...
CheeseMan42 Posted November 9, 2004 Posted November 9, 2004 actually there are only 2 roots, i have checked graphically and thanks to squarepi for the help Share this post Link to post Share on other sites More sharing options...
red930 Posted November 9, 2004 Posted November 9, 2004 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. Share this post Link to post Share on other sites More sharing options...
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