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discrimminant for quartics AS maths

please see attatched
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Reply 1
Avatar for joostan
joostan
13
Original post by CasualSoul
please see attatched


The quartic here is actually a quadratic in x2x^2 You could use a little substitution of u=x2u = x^2 to demonstrate this better if you wanted to :smile:
Reply 2
Avatar for kostis12345
kostis12345
I wouldn't really call it a quartic. I mean, yes, it is x^4 but the x^3 is missing therefore if you let b=x^2 you've got a nice quadratic on which you can use the discriminant.
About the second part, I'm not really sure. I would end it at k2+8>0 k^2+8>0 and say that this is true for any k therefore x4−kx2−2=0 x^4-kx^2-2=0 has real roots so y=x4−2y=x^4-2 and y=kx2y=kx^2 intersect for any k.
Reply 3
Avatar for CasualSoul
CasualSoul
OP
16
Original post by kostis12345
I wouldn't really call it a quartic. I mean, yes, it is x^4 but the x^3 is missing therefore if you let b=x^2 you've got a nice quadratic on which you can use the discriminant.
About the second part, I'm not really sure. I would end it at k2+8>0 k^2+8>0 and say that this is true for any k therefore x4−kx2−2=0 x^4-kx^2-2=0 has real roots so y=x4−2y=x^4-2 and y=kx2y=kx^2 intersect for any k.


thanks- gave you rep :d

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