Some strange math taking place here.
first of all, the way you typed it ...
4x^2+1/4x <1 , if you multiply by 4x you would get 16x^4 + 1 < 4x
According to your result, I suspect
that your question was:
(4x^2+1)/(4x) < 1
then 4x^2 + 1 < 4x+1
4x^2 - 4x + 1 < 0
(2x + 1)^2 < 0
which of course is not possible since the square of anything cannot be negative.
I have no idea what you did to get
x+1 < x+1
Prove that there exists no positive real number such that:
4x^2+1/4x <1
Can I simplify this inequality in the following way:
multiply both sides by 4x so we have.
4x^2+1 <4x+1
Which renders the original statement false as 4x^2+1 cant be less than 4x+1.
Even further can I divide both sides by 4x to leave the following:
x+1 < x+1
Which again proves the statement cant be true as its the same on either side.
Does this work? im thinking outside the box here.
4 answers
4x^2+1/4x <1
Can I simplify this inequality in the following way:
multiply both sides by 4x so we have.
4x^2+1 <4x+1
==========================
I have no idea what you have and what you are doing.
First of all, whatever, No
If you have
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x
you have
4 x^2 + 2 < 4 x
that means
4 x^2 -4x + 2 <0
graph the parabola, where is the vertex?
y = 2 x^2 -2x +1
x^2 - x = y-.5
x^2 - x + 1/4 = y -.25
(x-1/2)^2= y-.25
vertex at (1/2,1/4) ABOVE x axis
so no negative y values
the end
Can I simplify this inequality in the following way:
multiply both sides by 4x so we have.
4x^2+1 <4x+1
==========================
I have no idea what you have and what you are doing.
First of all, whatever, No
If you have
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x
you have
4 x^2 + 2 < 4 x
that means
4 x^2 -4x + 2 <0
graph the parabola, where is the vertex?
y = 2 x^2 -2x +1
x^2 - x = y-.5
x^2 - x + 1/4 = y -.25
(x-1/2)^2= y-.25
vertex at (1/2,1/4) ABOVE x axis
so no negative y values
the end
looks like both Damon and I have a typo
in mine: from 4x^2 - 4x + 1 < 0
(2x - 1)^2 < 0
the conclusion is still the same
in Damon's
from
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x
you would get:
4x^2 + 1 < 4x <---- which is the same as mine
in mine: from 4x^2 - 4x + 1 < 0
(2x - 1)^2 < 0
the conclusion is still the same
in Damon's
from
(4x^2+1)/4x <1 NOTE-parentheses VITAL
and multiply both sides by 4 x
you would get:
4x^2 + 1 < 4x <---- which is the same as mine
Yes, however you do it, the vertex is above the x axis :)