x^3 + 2x^2 - 1 = 0

Question

x^3 + 2x^2 - 1 >= 0

myininaya · Answer

$$f(x)=x^3+2x^2-1$$
$$f(-1)=0$$
so we have
-1 | 1    2     0       -1
     |      -1   -1        1
      -------------------------------------
       1     1   -1    |   0

so we have
$$f(x)=(x+1)(x^2+x-1)$$

saifoo.khan · Answer

synthetic division! :D

myininaya · Answer

$$f=0=> x=-1 $$ 
$$f=0=>x=\frac{-1 \pm \sqrt{1-4(1)(-1)}}{2}=\frac{-1 \pm \sqrt{5}}{2}$$

myininaya · Answer

so we have that we need to test the following intervals:
$$(-\infty,\frac{-1-\sqrt{5}}{2});(\frac{-1-\sqrt{5}}{2},-1);(-1,\frac{-1+\sqrt{5}}{2});(\frac{-1+\sqrt{5}}{2},\infty)$$

myininaya · Answer

test the first interval by plugging some number from that interval into f
and test the second interval by plugging some number from that interval into f
and so on

gina · Answer

$$x^3 +x^2+x^2 -1 >0 ...next i don't know .lolbfirst try \to expand (2x^2 into x^2 +x^2..then see where \it takes u..$$