Can someone help me solve this 16x^2> 9x... it is late and my brain is fried ;o)?

Answer 1

16x^2 > 9x
16x^2 - 9x > 0
x(16x - 9) > 0

Three ranges
x <= 0
0 < x < 9/16
x > 9/16

If x=-1
16x^2 = 16
9x = -9
16x^2 > 9x

Therefore
x < 0 ------------- 16x^2 > 9x
0 <= x <= 9/16 ------------- 16x^2 <= 9x
x > 9/16 ----------- 16x^2 > 9x

The required range is
x < 0 and x > 9/16
-----------------------------------------------------
Alternatively
f(x) = 16x^2 - 9x
2 zeros ---- x = 0 and x = 9/16
f'(x) = 32x-9
x = 9/32 which lies between 0 and 9/16

f"(x) = 32 ----- positive second derivative
It is a minimum
f(0) and f(9/16) = 0
f(9/32) < 0
Therefore f(x) is negative between 0 and 9/16
f(x) is positive for all values of x outside this range.

Answer 2

16x^2> 9x
16x > 9
x > 9/16 and x < 0 since putting any negative numbers would make the left positive and the right negative.

Answer 3

x > 9/16

Answer 4

16x^2 > 9x
16x^2 - 9x > 0
x^2 - (9/16)x > 0

x^2 - (9/16)x + 81/1024 - 81/1024 > 0

[ x - (9/32) ]^2 > 81/1024

| x - (9/32) | > 9/32

-9/32 < x - (9/32) < 9/32
0 < x < 18/32
0 < x < 9/16

Answer 5

Divide the expression by x to get

16x > 9 or x > 9/16

Done!

Answer 6

answer is a split function x<0 and x>9/16

my answer is correct. do the math. try x=-0.5. try x=8/16 and x=10/16