Can someone show me how to do this with steps please.... x^2-x-60

Question

Can someone show me how to do this with steps please.... x^2-x-6>0

anonymous · Answer

do you know how to factor quadratic equations?

anonymous · Answer

yes I factored it to (x+2)(x-3) but the book gave the answer as  x<-2 or x>3 ..why is that??

anonymous · Answer

because note that it is a quadratic inequality, not a quadratic equation.

anonymous · Answer

Yes i know it is inequality, but how does that change things, can you show me how it changes it please :)

anonymous · Answer

so this means that (x+2)(x-3)>0
now, suppose x = 3
your LHS becomes
(3+2)(3-3) = 5 (0) = 0
but we need a value of x which gives the LHS to be greater than RHS

anonymous · Answer

allow me to butt in. you have 
$$x^2-x-6>0$$ and you know ( i hope) that 
$$y=x^2-x-6$$ is a parabola that faces up. it has zeros at 
$$x=-2,x=3$$ so by the picture you know that it is positive on 
$$(-\infty,-2)\cup (3,\infty)$$ and negative on
$$(-2,3)$$ right from the picture.  positive means above the x - axis and negative means below.

anonymous · Answer

don't forget when you solve an inequality you know not only where something is greater than zero, but also where it is less than zero.  when you solve an inequality you solve for everything.

anonymous · Answer

so the "steps" are only to find the zeros.  it will be negative between then and positive outside of them

anonymous · Answer

:), Both explanations make sense to me, Thank you dhatraditya and satellite73 !!

anonymous · Answer

yw. btw here is a nice picture that shows you were it is positive and where it is negative http://www.wolframalpha.com/input/?i=y%3Dx^2-x-6

anonymous · Answer

Thanks, that is a wonderful website :)