Posted by Erin on Wednesday, December 10, 2008 at 4:13pm.
First of all, sketch a graph of
y = (x-2)(x-5)
This will be a parabola because it is a quadratic, a x^2 + b x + c
for large negative x, y is big positive
it is zero at x = 2
then it is negative for a while
it is then zero at x = 5
then it goes big positive
so
negative between x = 2 and x = 5
Thank you so much, Damon! You really helped me out! : )
I used to teach it with a slightly different approach because it worked well with multiple factors.
for example I will add another factor to your problem
(x-2)(x-5)(x+2) > 0
the x-intercepts of the corresponding function would be x = -2,2, and 5
mark those on the x-axis
there are now 4 sections to my x-axis
1. all values below -2
2. all values between -2 and 2
3. all values between 2 and 5
4. all values greater than 5
now pick an arbitrary value in each section and test it. You don't have to actually evaluate it, just worry about + or - (matching the > or <)
1. let x=-5 ---- (-)(-)(-) = - (does not work)
2. let x = 0 ---(-)(-)(+) = + works
3. let x = 3 ---(+)(-)(+) = - no
4. let x = 6 --- (+)(+)(+) = + yes
so -2 < x < 2 OR x > 5
for quadratic inequalities, I would also use the method outlined by Damon, but for cubics and higher, the above method works very nicely.
Of course if the function does not factor it would be a mess anyway.
Thank you, Reiny. My teacher taught me something very similar to your method and now I understand it. Thank you for your help! : )