# Math Help - solution set

1. ## solution 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

2. ## Re: solution set

1. If $x \ge 0$, then $|7x| = 7x$ and $7x = 4-x \Rightarrow x = \frac{1}{2}$.

If $x \le 0$ then $|7x| = -7x$, $-7x = 4-x \Rightarrow x = -\frac{2}{3}$. Both of these solutions satisfy the original equation.

2. Factor to $x(x-3)(x-5)(x+6) \le 0$. Exactly one or three of these factors must be positive. Use casework.

3. ## Re: 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

4. ## Re: solution set

Originally Posted by rcs
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
We know that \displaystyle \begin{align*} x(x - 3)(x - 5)(x + 6) \leq 0 \end{align*}. The LHS will be equal to 0 when \displaystyle \begin{align*} x = \{ -6, 0, 3, 5 \} \end{align*}. 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.

5. ## Re: solution set

will this have a very long solution? to find the solution set ?

6. ## Re: solution set

No, it's not long at all. There's a little brute-force involved, but not that much.

7. ## Re: solution set

thank you Guys... im trying but still i cannot figure out

8. ## Re: solution set

You want to solve $x(x-3)(x-5)(x-6) \le 0$. You know that if x = 0, 3, 5, or 6 you obtain equality. If $x < 0$, is the LHS positive or negative?

9. ## Re: 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?

10. ## Re: 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 $x(x-3)(x-5)(x-6)$ is positive. However we are only interested in the intervals where LHS is negative. Hence $x \le 0$ is not part of the solution set.

Repeat the same for $0 < x < 3$, $3 < x < 5$, $5 < x < 6$, and $6 < x$.

12. ## Re: solution set

I thought I explained this...

There are five important intervals you need to test, namely all the intervals around the roots, so \displaystyle \begin{align*} x < -6, -6 < x < 0, 0 < x < 3, 3 < x < 5, x> 5 \end{align*}.

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

\displaystyle \begin{align*} x(x - 3)(x - 5)(x + 6) &= -7(-7 - 3)(-7 - 5)(-7 + 6) \\ &= -7(-10)(-12)(-1) \\ &= 840 > 0 \end{align*}

Since this value is positive, then we can say that the region in which x < -6 does not satisfy \displaystyle \begin{align*} x(x - 3)(x - 5)(x + 6) \leq 0 \end{align*}.

Now try testing values of x from each of the other important regions. Accept the regions which give negative values for the function.

13. ## Re: solution set

Originally Posted by rcs
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?
I've always found that drawing a graph makes solving these inequalities easier.

The equation is: $x(x-3)(x-5)(x-6) \le 0$.

Go here: x&#40;x-3&#41;&#40;x-5&#41;&#40;x-6&#41; - 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 $x(x-3)(x-5)(x-6)$ is negative when $0 \le x \le3$

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