Solve quadratic inequality, graph, and write in interval notation?

Question

x^2 - 8x + 15 >0

Answer 1

x^2-8x-15>0
=(x-5)(x-3)>0
so x<3 and x>5
3>x>5
(-infinity,3)U(5,infinity)
graph will be a parabola

Answer 2

First: make an equation
x^2-8X+15=0
Then factorise:
(using the quadratic formula)
(-b+'square root' of b^2-4*a*c)/2a or
(-b-'square root' of b^2-4*a*c)/2a
with ax^2+-bx+-c=0
or this way (better in this case):
15=-3*-5 and -3-5=-8 --> (x-3)(x-5)=0
(x-5)=0 or (x-3)=0
5-5=0 or 3-3=0
y=0 when x=5 or when x=3
this means that
y>0 when x<3 and x>5

(chech this by plotting the graph)

I'm not sure about how to write an interval (i guess something like [3-5] or <3-5])

I hope you know a bit more now, but i find it difficult explaining things this way...

A useful site (in Dutch) first go to http://www.fi.uu.nl/wisweb/
than click the hyperlink DWO (in red)
click 'log in als gast' and go to 'kwadratische vergelijkingen' you can solve equations here, and by clicking 'eigen vergelijking' you can insert your own equations. I'm not sure whether this is useful for you, but I don't know any English?American sites...

Answer 3

Your first step to solve this is to factor the inequality.

x^2 - 8x + 15 > 0
(x - 5) (x - 3) > 0

x = 5 and x = 3 will be our key values in determining our intervals, because we have to investigate the area AROUND those values. They include the following intervals:

(-infinity, 3)
(3,5)
(5,infinity)

All we have to do is test one value that fall within those intervals. What we're looking for is a positive number (since we have a greater than or equal to sign).

First, let's test -100.
(x - 5) (x - 3) > 0
This gives us a negative number times a negative number, which is a positive number. Therefore, (-infinity, 3) is included in our solution.

Now, test a value between 3 and 5; let's test 4. Then
(x - 5) (x - 3) > 0
Gives us a negative times a positive, which is a negative. We discard the interval (3,5)

Test a value from 5 to infinity; let's test a million. We get:
positive times positive, which is positive. We include this interval.

Therefore, x lies in the interval (-infinity, 3) U (5,infinity)

Note that we use round brackets because, had we used square brackets for 3 and 5, we would be *including* zero. But the inequality in question precludes zero because it is strictly greater than (>) as opposed to greater than or equal to (>=) 0.

Answer 4

you got to do your homework yourself. From just getting the solution, you don't learn anything.