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: \( (\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} ) \)
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: \( (\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} ) \)
48x - 16x^2 > 30 rewrite as
0 > 16x^2 - 48x + 30 rewrite again as
16x^2 - 48x + 30 < 0 divide through by 2
8x^2 - 24x + 15 < 0 set equal to 0
8x^2 - 24x + 15 = 0
Using the quadratic formula, we have
x = [24 ± √[ (-24^2 - 4(8)(15) / [ 2 * 8] =
[ 24 ± √96] / 16 = [ 24 ± 4√6] / 16 = [ 6 ± √6] / 4
We have 3 intervals to test here......either the middle interval works or the two "outside" intervals do
The intervals are (-inf, [6 - √6]/ 4 ) , ( [6 - √6]/ 4, [6 +√6]/ 4 ), ( [6 + √6]/ 4 , inf)
Pick a point in the middle interval - I'll choose 1 - and test it in the original problem
48(1) - 16(1)^2 > 30
32 > 30 this is utrue......thus this interval solves the problem
And the solution is : ( [6 - √6]/ 4, [6 +√6]/ 4 )