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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- Oct 6th 2009, 07:41 PMagent2421Math 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. - Oct 6th 2009, 07:51 PMartvandalay11
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,

So the inequality fails

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

If x=-2,

So the inequality holds

Let's try 1 for 0<x<5

If x=1,

So the inequality fails

Let's try 10 for 5<x

If x=10,

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) - Oct 7th 2009, 01:28 AMGrandad
Hello everyone!

Just a slight variation on artvandalay11's method is to express as factors:

before plugging your 'test' values of .

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

gives , which is because .

Grandad