can anyone tell me what solution i could use to find the solution set of:

1.) |7x| = 4 - x

2. (x^2 - 8x + 15) ( x^2 + 6x) <= 0

thank you

Printable View

- July 30th 2012, 10:22 AMrcssolution set
can anyone tell me what solution i could use to find the solution set of:

1.) |7x| = 4 - x

2. (x^2 - 8x + 15) ( x^2 + 6x) <= 0

thank you - July 30th 2012, 10:25 AMrichard1234Re: solution set
1. If , then and .

If then , . Both of these solutions satisfy the original equation.

2. Factor to . Exactly one or three of these factors must be positive. Use casework. - July 30th 2012, 06:26 PMrcsRe: solution set
by cases : is it possible to say that (x-3)>= 0 , or (x-5)> = 0, or (x+6)>= 0, what about x ? is that x > =0

is that how i am going to solve for the values of x or roots sir?

thank you - July 30th 2012, 07:52 PMProve ItRe: solution set
We know that . The LHS will be equal to 0 when . The only times that a function will go from negative to positive or vice versa is when it crosses the x axis, so at these values of x. So test values of x on each side of these intercepts to see where the function is negative.

- July 30th 2012, 09:22 PMrcsRe: solution set
will this have a very long solution? to find the solution set ?

- July 30th 2012, 09:29 PMrichard1234Re: solution set
No, it's not long at all. There's a little brute-force involved, but not that much.

- July 30th 2012, 09:45 PMrcsRe: solution set
thank you Guys... im trying but still i cannot figure out :(

- July 30th 2012, 09:59 PMrichard1234Re: solution set
You want to solve . You know that if x = 0, 3, 5, or 6 you obtain equality. If , is the LHS positive or negative?

- July 30th 2012, 10:25 PMrcsRe: solution set
do you mean sir that i have to make this - x <= 0 then x <= o : -x - 3<= 0 ; - x - 5 <= 0 : - x - 6 <= 0 .... so solve for x?

- July 30th 2012, 10:39 PMrichard1234Re: solution set
Sort of, but you're not solving for x. If x is negative, then x-3, x-5, and x-6 are negative, and the product is positive. However we are only interested in the intervals where LHS is negative. Hence is not part of the solution set.

Repeat the same for , , , and . - July 30th 2012, 10:58 PMrcsRe: solution set
huhuhuhuh :( what can i do sir ... my brain is not working anymore please help

- July 31st 2012, 01:30 AMProve ItRe: solution set
I thought I explained this...

There are five important intervals you need to test, namely all the intervals around the roots, so .

So choose a value of x in the first interval, say x = -7. Then we have

Since this value is positive, then we can say that the region in which x < -6 does not satisfy .

Now try testing values of x from each of the other important regions. Accept the regions which give negative values for the function. - July 31st 2012, 01:36 AMjgv115Re: solution set
I've always found that drawing a graph makes solving these inequalities easier.

The equation is: .

Go here: x(x-3)(x-5)(x-6) - Wolfram|Alpha to look at your polynomial.

Now found the values of x that make the graph go below the x-axis - as that's essentially what the equation is saying.

So looking at the graph, it's obvious that values of is negative when

Do you see how I got that? Now you find the next solution