Hello,
Think "discriminant", in .
If , there is no zero within the real numbers.
If , there is one zero.
If , there are 2 zeroes.
Hello,
Think "discriminant", in .
If , there is no zero within the real numbers.
If , there is one zero.
If , there are 2 zeroes.
No, there are the values where (x+2)(x-8)<0, so that will be sets of values.
I'll give you a hint : a product of two factors is negative if and only if the two are of different sign.
This means that you have to solve two things :
- x+2>0 AND x-8<0
- x+2<0 AND x-8>0
One of the situation will not be possible
You can do it in two ways
The one I tried to describe with solving k+2>0 etc...
And another one, which is simpler, that's why I changed the method :
The roots are -2 and 8.
I don't know if you have learnt that if a polynomial has roots and (and ), then we can have the sign of the polynomial :
- sign of a if
- opposit sign of a if
Here, a=1, what can you conclude ?
You have:
So we have the points and .
is positive so the part to the right of is positive, the part to the left of will also be positive.
We want the part smaller than .
Those numbers lie between and , but the two aforementioned numbers are not included.