SOLUTION: Please help me solve this: {{{2x^3 - x^2 >= 0}}} <=> {{{x^2*(2x - 1) >= 0}}} and solutions are: x = 0 or x >= 1/2, because: {{{2x - 1 >= 0}}} <=> {{{x >= 1/2}}} {{{x^2 >= 0}}}

Algebra ->  Inequalities -> SOLUTION: Please help me solve this: {{{2x^3 - x^2 >= 0}}} <=> {{{x^2*(2x - 1) >= 0}}} and solutions are: x = 0 or x >= 1/2, because: {{{2x - 1 >= 0}}} <=> {{{x >= 1/2}}} {{{x^2 >= 0}}}       Log On


   

Question 703877: Please help me solve this:
2x%5E3+-+x%5E2+%3E=+0 <=> x%5E2%2A%282x+-+1%29+%3E=+0
and solutions are: x = 0 or x >= 1/2, because:
2x+-+1+%3E=+0 <=> x+%3E=+1%2F2
x%5E2+%3E=+0 <=> x+%3E=+sqrt%280%29 <=> x+%3E=+0
but why x = 0 and not x >= 0?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
AN PICTURE IS WORD A THOUSAND WORDS:
This is how my 11th grade math teacher taught me how to solve that kind of problem. (Teachers may have tried to teach me some math earlier than that, but I was not paying too much attention).
The sign of a product:
You place pluses, minuses, and zeros on the stacked (lined up) number lines for each factor.
Then, you figure out the pluses, minuses, and zeros of the product and write them on the bottom number line (the one for the product)

GRAPHS:
x%5E2 graph%28300%2C300%2C-1.5%2C1.5%2C-0.5%2C0.5%2Cx%5E2%29 times %282x-1%29 graph%28300%2C300%2C-2%2C2%2C-5%2C5%2C2x-1%29 equals x%5E2%282x-1%29 graph%28300%2C300%2C-1.5%2C1.5%2C-0.5%2C0.5%2Cx%5E2%282x-1%29%29

AN EXPLANATION IN MANY WORDS:
x%5E2 is zero for x=0
For all other values of x, x%5E2%3E0 is positive,
and x%5E2%2A%282x+-+1%29+ will go along with %282x-1%29,
being negative for all x%3C1%2F2,
except for x=0, where %282x-1%29%3C0,
but because x%5E2=0, x%5E2%282x-1%29=0
At that point x%5E2%282x-1%29 refuses to go along with %282x-1%29.