# Math Help - Help me check this quadratic equation

1. ## Help me check this quadratic equation

$\frac{x^2+1}{x}+\frac{x}{x^2+1}=2.9$
Substitution: $z=\frac{x^2+1}{x}$
$z+\frac{1}{z}=2.9$
$z^2-2.9z+1=0$
$z_1,_2=\frac{2.9\mp\sqrt{4.41}}{2}$
$z_1=2.5$
$z_2=0.4$

So

$2.5=\frac{x^2+1}{x}$
$x_1,_2=\frac{2.5\mp\sqrt{2.25}}{2}$
$x_1=2$
$x_2=0.5$

and

$0.4=\frac{x^2+1}{x}$
$x^2-0.4x+1=0$
$x_3,_4=\frac{0.4\mp\sqrt{0.16-4}}{2}$

Which brings us to no solution

Is this right?

2. Originally Posted by nathan02079
$\frac{x^2+1}{x}+\frac{x}{x^2+1}=2.9$
Substitution: $z=\frac{x^2+1}{x}$
$z+\frac{1}{z}=2.9$
$z^2-2.9z+1=0$
$z_1,_2=\frac{2.9\mp\sqrt{4.41}}{2}$
$z_1=2.5$
$z_2=0.4$

So

$2.5=\frac{x^2+1}{x}$
$x_1,_2=\frac{2.5\mp\sqrt{2.25}}{2}$
$x_1=2$
$x_2=0.5$

and

$0.4=\frac{x^2+1}{x}$
$x^2-0.4x+1=0$
$x_3,_4=\frac{0.4\mp\sqrt{0.16-4}}{2}$

Which brings us to no solution

Is this right?
i did not work it out, but i believe it has a solution. your $z_2$ should be ${\color {red} -} 0.4$

EDIT: in fact, the solutions are x = 1/2 and x = 2

3. Originally Posted by Jhevon
i did not work it out, but i believe it has a solution. your $z_2$ should be ${\color {red} -} 0.4$

EDIT: in fact, the solutions are x = 1/2 and x = 2
but isnt
$z_2=\frac{2.9-2.1}{2}$?
then
$z_2=\frac{0.8}{2}$
which is then
$z_2=0.4$

well at least thats how i came up with the answer:P

4. Originally Posted by nathan02079
but isnt
$z_2=\frac{2.9-2.1}{2}$?
then
$z_2=\frac{0.8}{2}$
which is then
$z_2=0.4$

well at least thats how i came up with the answer:P
yes. you are correct, my bad. i did 2.1 - 2.9 instead of 2.9 - 2.1

so you made a mistake somewhere else then... let's see...ah, no mistake. you got the two solutions i gave! x_1,2 is your answer. the others are complex simply because no (real) solution exists for that value of z. it's fine