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Math Help - Math Question Help

  1. #1
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    Math Question Help

    Solve for x, ER

    -x^3+25x < 0

    ----

    I tried it out and i got:

    x = 0, -5, 5

    therefore if x is less than 5, x < 0
    if x is greater than 5, x <0

    not sure if it's right though.
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  2. #2
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    Quote Originally Posted by agent2421 View Post
    Solve for x, ER

    -x^3+25x < 0

    ----

    I tried it out and i got:

    x = 0, -5, 5

    therefore if x is less than 5, x < 0
    if x is greater than 5, x <0

    not sure if it's right though.
    You are very close, but yes there is something wrong

    You succeeded in finding the points at which the expression=0, now what we do is plug values in between those points into the expression to see if the inequality is satisified

    So we have -5,0, and 5

    If you draw a number line and plot these points on it, you will see the number line gets divided into 4 regions namely:

    x<-5
    -5<x<0
    0<x<5
    5<x

    So we have to test values in each of those regions

    So let's try -10 for x<-5,

    If x=-10, -x^3+25x=-(-10)^3+25(-10)=-(-1000)-250=1000-250=750>0
    So the inequality fails

    Now let's try -1 for -5<x<0

    If x=-2, -x^3+25x=-(-1)^3+25(-1)=-(-1)-25=1-25=-24<0
    So the inequality holds

    Let's try 1 for 0<x<5

    If x=1, -x^3+25x=-(1)^3+25(1)=-(1)+25=-1+25=24>0
    So the inequality fails

    Let's try 10 for 5<x

    If x=10, -x^3+25x=-(10)^3+25(10)=-(1000)+250=-1000+250=-750
    So the inequality holds

    So the answer is where the inequality holds, namely:

    -5<x<0
    x>5

    When stating your answer you want to say

    -5<x<0 or x>5

    (A lot of people would say "And" not "or" but this is wrong since x cannot be both <0 and >5)

    Also note that it doesn't matter what numbers to choose in the intervals (so choose numbers that are easy to work with)
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  3. #3
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    Hello everyone!

    Just a slight variation on artvandalay11's method is to express -x^3+25x as factors:

    -x^3+25x = x(-x^2+25)

    =x(5+x)(5-x)

    before plugging your 'test' values of x.

    It makes the arithmetic much easier, because you don't need to work out the actual value, just its sign. For instance:

    x=-10 gives (-10)\times(-5)\times(15), which is >0 because (-^{ve})\times(-^{ve})\times(+^{ve})=(+^{ve})
    .

    Grandad
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