Solve the equation x^3 + 3ax^2 - 5a^2x?
回答 (5)
2009-05-04 8:54 am
✔ 最佳答案
x³ + 3ax² - 5a²x = 0x(x² + 3ax - 5a²) = 0
x=0 or x²+3ax-5a²=0 ...use the quadratic formula:
x = 0, (-3+√29)/2, -a(3+√29)/2
2009-05-04 4:32 pm
Actually, this is an expression.
x^3 + 3ax^2 - 5a^2x
= x(x^2 + 3ax - 5a^2)
x^3 + 3ax^2 - 5a^2x
= x(x^2 + 3ax - 5a^2)
2009-05-04 3:56 pm
for x^3 + 3ax^2 - 5a^2x = 0
divide the whole equation by x
u get x^2 + 3ax - 5a^2 = 0
u can solve this by any of the following methods
1) factorization method
2) quadratic formula
yeilds
x = (-3a +/- Sq.root 29a^2)/2
plz don't forget there is "a" also which will have some value.
divide the whole equation by x
u get x^2 + 3ax - 5a^2 = 0
u can solve this by any of the following methods
1) factorization method
2) quadratic formula
yeilds
x = (-3a +/- Sq.root 29a^2)/2
plz don't forget there is "a" also which will have some value.
2009-05-07 9:55 pm
=x(x^2+3ax-5a^2)
2009-05-04 3:51 pm
Well, you can factor it. The only way I can do it is by guess and check, but you can see each term has an X in it, so factor that out first:
x ( x^2 + 3ax - 5a^2)
This does not factor evenly any further. So this could be your answer.
The fact is says "solve" confuses me too. From here you could use the quadratic formula if the equation was equal to something. But nothing was mentioned, so that's as far as I can solve for.
x ( x^2 + 3ax - 5a^2)
This does not factor evenly any further. So this could be your answer.
The fact is says "solve" confuses me too. From here you could use the quadratic formula if the equation was equal to something. But nothing was mentioned, so that's as far as I can solve for.
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