Find the range of values of k if kx^2+8x>6-k for all real values of x
What does for all the real values of x mean?
The answer is k>8 and I can't seem to get it
this is my working by the way:
Kx^2 +8x+k-6 >0
a=k b=8 c=k-6
sub into b^2-4ac
4k^2-24k-64<0
k^2-6k-16<0
(k-8)(k+2)<0
thank you!
Notice you have the product of two things being positive. This can only happen if both things are positive or both things are negative.
Case 1: and . Working on the second inequality gives
Now clearly this can only have solutions if , so
So putting the inequalities together, we find that .
Now try Case 2, the situation where both quantities have to be negative...
For the quadratic equation we do call is the discriminant.
The name comes from the fact it discriminates between real and complex roots.
If the quadratic has no real roots. That means the entire graph is either above the x-axis or below the x-axis.
Let's review the OP
Here . If we see so there are no real roots.
So this leaves one possible solution set: