Question

solve the following inequality. Write the answer in interval notation. -x^2+6x>0

Answer 1

solve it like \[x ^{2}+6x =0\]
and then put the Greater than sign back in

Answer 2

\[-x^2 + 6x > 0\]\[\Rightarrow -(x^2 - 6x) > 0\]\[\Rightarrow x^2 - 6x < 0\]Do you follow till here? Now comes the interesting part...

Answer 3

We can write \(x^2 - 6x\) as \(x(x -6)\). This means that we have it equal to zero at \(x=0\) or \(x = 6\).

Answer 4

Now, let's try to see what happens between \(x = 0\) and \(6\). Take \(x = 3\), for example. \((3)^2 - 6(3) = -9\). So, we can somehow say that \(x^2 - 6x<0\) for \(x \in (0,6)\)

Answer 5

That was great! thanks!