Question

if 2x^2+1>3x, prove that |4x-3|>1

actually i ald has d ans but i don't know is right or wrong

Answer 1

here is the ans:
2x^2+1>3x
2x^2-3x+1>0
(2x-1)(x-1)>0
i solve by graph,
x<1/2 ------1st
or
x >1--------2nd

use 1st inequality
x<1/2
2x<1
4x<2
4x-3<2-3
4x-3<-1
-(4x-3)>1

use the 2nd inequality
x>1
4x>4
4x-3>4-3
4x-3>1

by combine the inequalities -(4x-3)>1 and 4x-3>1,
Therefore,
|4x-3|>1

Answer 2

so my ans correct??

Answer 3

i don't know but you typed a lot i gave you medal



let me check

Answer 4

its right...

Answer 5

Okay let me do this

\[2x^2 -3x + 1>0\]
\[(2x-1)(x-1)>0\]


1 1/2
---------|---------------|-----------
++++ -------- ++++++

Okay you did that right

|4x-3|>1
-1>4x-1 or 4x-3>1
-1/2>x or x>1

Answer 6

ya but until the number line.... the lower part u can refer to the 1st post.... i write there...

Answer 7

neoh my head not working now i gotta sleep ..what i did is correct if your book shows something other then it's wrong

Answer 8

x<-1/2 or x>1 while x is in quadratic one x >1/2 or x<1 just check the domains

Answer 9

ok... my book has ans only, no step.... especially proving question(no ans).... but thanks a lot for helping me...

Answer 10

good luck : ) i will get some help here

Answer 11

neoh, you did it correctly... what's the question?
The only thing I can add is that you can solve the original using algebra rather than graphically:
(2x-1)(x-1)>0 is true if

1) both terms are positive. i.e. 2x-1> 0 and x-1>0
this gives x>1/2 and x>1. since both terms must be positive, this simplifies to x>1
2) both terms are negative. i.e. 2x-1<0 and x-1<0
this gives x<1/2 and x<1. Both conditions must be true, so x<1/2 (Note that if x <1 this would allow for example x= 3/4, which is > 1/2, so the 2x-1 term would not be negative).

Everything you did with these two inequalities makes perfect sense.

Answer 12

ok... it's good... my teacher didn't teach me this method... it is a good method... thx...:-)