Question

solve the inequality for x

x^3-5x greater then or equal to 4x^2
i simplified to x(x+5)(x-1) but its saying that is incorrect

Answer 1

\[x^{3}-5x \ge 4x ^{2}\]

Answer 2

I'm solving hold on

Answer 3

i figured it out and its \[x(x-5)(x+1)\] how do i figure out which are included in the solution set

Answer 4

\[x^3-5x\geq 4x^2\]

\[x^3-4x^2-5x\geq0\]
\[x(x^2-4x-5)\geq 0\]
\[x(x+4)(x-5)\geq0\]
zeros are at -4, 0 and 5and since this is a cubic polynomial with positive leading coefficient, it will be negative, then positive, then negative, then positive (think of the picture)
you want it to be positive, so answer is
\[[-4,0]\cup [5,\infty)\]

Answer 5

in your head draw this picture

Answer 6

it says the answer is (-1,0] U [5, infinity)

Answer 7

because if you factor out (x+4)(x-5) you get \[x ^{2}-1x-30\] which isnt what i want

Answer 8

solution
x>=5
-1<=x<=0