30 < 48x - 16x2
0 < -16x2 + 48x - 30
Let's set this = 0 and then use the quadratic formula to solve for x.
\(x = {-48 \pm \sqrt{48^2-4(-16)(-30)} \over 2(-16)}={-48 \pm \sqrt{2304-1920} \over -32}={-48 \pm 8\sqrt{6} \over -32}=\frac{-6\pm \sqrt6}{-4} \\~\\ x=\frac{-6+\sqrt6}{-4} = \frac{6-\sqrt6}{4} \hspace{2cm}\text{or}\hspace{2cm} x=\frac{-6-\sqrt6}{-4}=\frac{6+\sqrt6}{4}\)
So -16x2 + 48x - 30 = 0 when x = the above values.
But we want to know what x values cause it to be greater than 0.
We want to know what x values make this true: 0 < -16x2 + 48x - 30
It will either be x values in the interval: \((\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} )\)
OR it will be x values in the interval: \( (-\infty, \frac{6-\sqrt6}{4} ) \cup (\frac{6+\sqrt6}{4} , \infty ) \)
To determine which, test a number for x and see if it makes a true statement.
Let's test x = 0, which is in the second interval listed.
0 < -16(0)2 + 48(0) - 30
0 < -30 false
Let's test x = 1, which is in the first interval listed.
0 < -16(1)2 + 48(1) - 30
0 < 2 true
Therefore x is in the interval: \( (\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} ) \)
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