Question

How do you solve 9 - 16x² > 0?

Answer 1

you can see that this is in the form of a^2-b^2=(a-b)(a+b)
that is in this case
3^2-(4x)^2=(3-4x)(3+4x)>0
when will this be greater than 0?

Answer 2

when both are either positive or negative, that is

Answer 3

yes! So what you need to do is solve 3-4x>0 and 3+4x>0
plus
0>3-4x and 0>3+45

Answer 4

-16x^2>9
16x^2<-9
divide both sides by 16
x<-9/16

Answer 5

?

Answer 6

no not right

Answer 7

if you graph
\[y=9-16x^2\]
you have a parabola that is facing down. it is zero at
\[-\frac{3}{4}\] and
\[\frac{3}{4}\]

Answer 8

it will be positive (above the x - axis) between the zeros and negative otherwise

Answer 9

16 x^2<9
(x-3\/4) (x+3\/4)>0
-(4 x-3) (4 x+3)>0
-3/4<x<3/4

Answer 10

you want to know where it is positive, so answer is ... what godefroid wrote : \[(-\frac{3}{4},\frac{3}{4})\]

Answer 11

Thank you so much :)!