thx but for the first one how'd you know it = (x^2+2x-4)(x^2+2x+5)=0 on the first step??
So or
For the other one, the double root will be a root of the derivative, so and .
So
So
answer that one please xDDIt's a bit boring though.
See if you can solve exactly.
(The 1969 3 unit paper asked for the approximate solution to this using Newton's method, but you can actually solve it exactly.)
I too would like to know!It's a bit boring though.
See if you can solve exactly.
(The 1969 3 unit paper asked for the approximate solution to this using Newton's method, but you can actually solve it exactly.)
have you tried getting a Girl friend to fill in your spare time?Somewhat uglier, but equivalent expressions from wolframalpha.com yield the four roots, the last two of which are the complex roots:
Now we can go back to fullonoob's question, modify it a bit and get more interesting ones by just removing some terms. This actually makes the question much harder and much more interesting:Solve the equation x^4 + 4x^3 + 5x^2 + 2x - 20 = 0