How do you solve $9x^2+3x-2>=0$ ?

Question

3979 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-26T19:17:32

$x=1/3>0$

$9x^2+3x-2-<0$

$:.=(3x-1)(3x+2)=0$

$:.3x-1=0$

$:.3x=1$

$:.x=1/3$

sustitute $x=1/3$

$:.9(1/3)^2+3(1/3)-2=0$

$:.9 xx 1/9+3(1/3)-2=0$

$:.9/9+3/3-2=0$

$:.1+1-2=0$

$3x+2=0$

$:.3x=-2$

$:.x=-2/3$

substitute $x=-2/3$

$:.9(-2/3)^2+3(-2/3)-2$

$:.9(-4/9)+3(-2/3)-2$

$:.-36/9-6/3-2=0$

$:.-4-2-2=0$

$:.-8=0$ an extraneous solution

How do you solve 9x^2+3x-2>=09x^2+3x-2>=0?