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Math Help - solution set

  1. #1
    rcs
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    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
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    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.
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    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?

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    Re: solution set

    Quote Originally Posted by rcs View Post
    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.
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    Re: solution set

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

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

    thank you Guys... im trying but still i cannot figure out
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    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?
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    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?
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    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.
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    Re: solution set

    huhuhuhuh what can i do sir ... my brain is not working anymore please help
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    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.
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    Re: solution set

    Quote Originally Posted by rcs View Post
    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
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