Question 44814
{{{(x-5)(x+2)<0}}}
Find the values where (x-5)(x+2)=0.
The answer is when x=-2 or x=5.
If you expand the brackets using FOIL; you get:
{{{x^2-3x-10<0}}}
in a qaudratic equation, if the amount of {{{x^2}}} is positive (>0), the graph of that function has a 'smiley' shape. if the amount of x-squared's is negative (<0), the graph has a 'frown' shape:
{{{x^2+x+1}}} is happy:
{{{ graph( 300, 200, -6, 5, -2, 10, x^2+x+1) }}} 
{{{-x^2+x+1}}} is frowning:
{{{ graph( 300, 200, -6, 5, -10, 2, -x^2+x+1) }}}
If we condider the graph of our function {{{y=x^2-3x-10}}}, we can see from the amount of {{{x^2}}}'s that it is a 'happy' graph. The value of y is greater as we move out from the minimum point of the graph. We know that y=0 when x=-2 and x=5, and because of it's 'happy' shape, we can deduce that the function is <0 between these values of x.
So {{{x^2-3x-10<0}}} when -2<x<5.
Check this with the graph of {{{y=x^2-3x-10}}}:
{{{ graph( 300, 200, -4, 10, -12, 5, x^2-3x-10) }}}
I hope this helps.
P.S. In am currently constructing my own online tutoring website. I would be extremely grateful if you would e-mail me some feedback on the help you received to adam.chapman@student.manchester.ac.uk
Adam