I can't tell what the question is
Original post by CorpusLuteum
I can't tell what the question is
its at the top
Original post by p29
its at the top
Is the quadratic equal to zero.
Are you only finding x? Why is there a Y.
Be more clear.
Original post by p29
2x^4+14x^2+24
I got one solution but I don't know why the other solution is supposed to be 2(x^2+3) when I divided the equation by 2 at the beginning please help.
I got one solution but I don't know why the other solution is supposed to be 2(x^2+3) when I divided the equation by 2 at the beginning please help.
As above, be more clear.
Are you solving ??? Or are you rearranging for something?
You let got up to which is almost correct.
So or
Hence or
Which gives you complex roots.
(edited 7 years ago)
Original post by p29
its at the top
You're overcomplicating it - the solution should only really be three lines max.
Start with the equation, divide by 2, factorise it (as of course) and then solve. Just because the quadratic is in as opposed to doesn't mean you can't solve it in the normal way.
Original post by RDKGames
As above, be more clear.
Are you solving ??? Or are you rearranging for something?
You got up to
So or
Hence or
Which gives you complex roots.
Are you solving ??? Or are you rearranging for something?
You got up to
So or
Hence or
Which gives you complex roots.
=0 and I have to factorise it
Original post by p29
=0 and I have to factorise it
If you have to factorise it then you've already done it with the y's, so replace them with x2 and you're done.
Original post by p29
=0 and I have to factorise it
the book says that when it is factorised one of them will be 2(x^2+3) which i dont understand
(edited 7 years ago)
Okay what factorisation have you got so far?
Original post by p29
is the book wrong?
No. You have factored out a 2 at the start so you need to bring it back.
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