# Thread: Basic Inequality

1. ## Basic Inequality

I have (x+8)(x+1) < 3x

I know that x^2 + 9x + 8 < 3x

But what happens after that?

2. after this you get

$(x+2)(x+4)<0$

$
\Rightarrow-4$

3. Yes, you first step is rite.

Inequalities can be confusing, but with most you can make it easier by imagining the > < or signs are an =, as unless you are multiplying by a negative number at any point then they behave the same.

To make it easier here we'll use it:

x^2 + 9x + 8 = 3x

that might not look as daunting, well first of all, as you know , we are trying to solve the equation so we cant have 3x as our answer, so we take it over to the other side:

x^2 +6x + 8 = 0

then we just have a 'quadratic' this can be solved multiple ways, luckily this can be solved the easiest way, 'factorising' (if you want more help on that message me) this factorises to:

(x + 4)(x + 2) = 0

(if you multiply out the brackets you'll get the original equation)

As we are multiplying 2 thing together and the product is 0, then one of them must be 0, so either could be:

If (x + 4)=0 then x=-4

If (x + 2) = 0 then x=-2

IMPORTANT: Remember to switch the = sign back to the <, otherwise the answer is wrong.

So your answers are x>-4 and x<-2

The reason its x>-4 is that its completely illogical if you had 'anything less than -4 and 'anything less than -2' because anything less than '4 is less than -2 already so we try the other side and get 'anything greater than 4'.

Also its better to express your answer in the 'proper' way, which is
-4 < x < -2

I hope this helped.