F fresh83 New member Joined Apr 17, 2009 Messages 20 Apr 29, 2009 #1 ummmm im doing some problems and i came across the equation x^4-5x^2+4=0 i know its going to have 4 answers ....but i dont remember how or how to get 4 answers ...the way im doing it is only turning up 2
ummmm im doing some problems and i came across the equation x^4-5x^2+4=0 i know its going to have 4 answers ....but i dont remember how or how to get 4 answers ...the way im doing it is only turning up 2
F fresh83 New member Joined Apr 17, 2009 Messages 20 Apr 29, 2009 #2 Re: quadradics with 4 answers and after making this thread im remembering something about setting x^2 = w . just letting everyone know what i know or....dont .
Re: quadradics with 4 answers and after making this thread im remembering something about setting x^2 = w . just letting everyone know what i know or....dont .
L Loren Senior Member Joined Aug 28, 2007 Messages 1,298 Apr 29, 2009 #3 Re: quadradics with 4 answers x^4-5x^2+4=0 (x^2-1)(x^2-4)=0 Now set each of these factors equal to 0 and solve for x. You should get 2 roots for each for a total of 4 roots. As a reminder... Solve x^2-9=0 Two ways. x^2-9=0 x^2=9 \(\displaystyle \sqrt{x^2}=\pm \sqrt{9}\) \(\displaystyle x=\pm 3\) Second way. x^2-9=0 (x+3)(x-3)=0 x+3=0 or x-3=0 x=-3, x=3.
Re: quadradics with 4 answers x^4-5x^2+4=0 (x^2-1)(x^2-4)=0 Now set each of these factors equal to 0 and solve for x. You should get 2 roots for each for a total of 4 roots. As a reminder... Solve x^2-9=0 Two ways. x^2-9=0 x^2=9 \(\displaystyle \sqrt{x^2}=\pm \sqrt{9}\) \(\displaystyle x=\pm 3\) Second way. x^2-9=0 (x+3)(x-3)=0 x+3=0 or x-3=0 x=-3, x=3.
