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# question

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what number does x equal if 30 is less than 48x - 16x2

Guest Apr 18, 2017

#1
+4154
+2

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: $$(\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} )$$

hectictar  Apr 18, 2017
edited by hectictar  Apr 18, 2017
Sort:

#1
+4154
+2

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: $$(\frac{6-\sqrt6}{4} , \frac{6+\sqrt6}{4} )$$

hectictar  Apr 18, 2017
edited by hectictar  Apr 18, 2017
#2
+75302
+2

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 )

CPhill  Apr 18, 2017

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