SOLUTION: its all about non linear inequalities. (x-5)(x-4) is greater than or equal to 0.

Question 476572: its all about non linear inequalities. (x-5)(x-4) is greater than or equal to 0.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
easiest way to answer this is to graph it.

that's if you have the graphing software that makes it easy.
from the graph, it looks like this equation will be less than 0 between 4 and 5.
that stands to reason.
since this is a quadratic equation that has already been solved for the roots, it should be easy to determine what points in this equation are above 0 or below 0 or at 0.
the equation you have to work with is:
(x-5)*(x-4) = 0
by setting the expression equal to 0, you are looking for the roots.
the roots becomes:
x = 4 and x = 5
those are the zero points of the graph.
you multiply the factors together to get the original quadratic equation that gave you (x-5)*(x-4.
(x-5)*(x-4) = x^2 - 4x -5x + 20
combine like terms and you get:
x^2 - 9x + 20
if you set this equal to 0, then it becomes the standard form of the quadratic equation.
you get x^2 - 9x + 20 = 0
the standard form of the quadratic equation is ax^2 + bx + c = 0
this makes:
a = 1
b = -9
c = 20
the min/max point of the quadratic equation is given by the equation:
x = -b/2a
this becomes:
x = -(-9)/2) which becomes:
x = 4.5
That's the x value of the min/max point.
the y value of the min/max point is given by:
y = f(4.5)
you replace x with 4.5 in the equation and you get:
y = x^2 -9x + 20 which becomes:
y = (4.5)^2 -9*4.5 + 20 which becomes:
y = -.-025
your min/max point is at the (x,y) coordinates of (4.5,-.025).
now you want to determine whether this is a min point or a max point.
this is because you didn't see the graph i just showed you above.
you haven't even drawn the graph yet.
you're working just from the equation.
you look at the exponent of the x^2 term.
if it is positive then the graph points down and opens up.
if it is negative then the graph points up and opens down.
if the graph points down, then the min/max point is a min point.
if the graph points up, then the min/max point is a max point.
the graph is pointing down because the coefficient of the x^2 term is positive.
that would be the a in ax^2 which was equal to 1.
so your graph is point down.
the roots of the eqution are at x = 4 and x = 5
the min point is negative.
it will stay negative between the roots and it will be positive outside of the roots.
your equation is therefore >= 0 when:
x < 4 or x > 5

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