# [SOLVED] Help on Quadratic Equation

• Sep 2nd 2009, 06:44 PM
sleels
Hello, I need some help with this problem.

The problem is:

$\displaystyle Ln(x) + Ln(x-1) = 1$

Heres what I got
----------------

$\displaystyle Ln(x) + Ln(x-1) = 1]$

$\displaystyle Ln(x^2 - x) = 1$

$\displaystyle x^2 - x = e^1$

$\displaystyle x^2 - x - e = 0$

Now I use the quadratic equation to get

$\displaystyle \frac{x \pm \sqrt{x^2 - 4(x^2)(-e)}}{2x^2}$

The answer is supposed to be :
$\displaystyle \frac{(1 \pm \sqrt{1 + 4e})}{2}$

I'm having some difficulty seeing how they got that answer.

Did I do something incorrectly? Or am I simplifying wrong?

Thank you
• Sep 2nd 2009, 08:01 PM
mr fantastic
Quote:

Originally Posted by sleels
Hello, I need some help with this problem.

The problem is:

$\displaystyle Ln(x) + Ln(x-1) = 1$

Heres what I got
----------------

$\displaystyle Ln(x) + Ln(x-1) = 1]$

$\displaystyle Ln(x^2 - x) = 1$

$\displaystyle x^2 - x = e^1$

$\displaystyle x^2 - x - e = 0$

Now I use the quadratic equation to get

$\displaystyle \frac{x \pm \sqrt{x^2 - 4(x^2)(-e)}}{2x^2}$

The answer is supposed to be :
$\displaystyle \frac{(1 \pm \sqrt{1 + 4e})}{2}$

I'm having some difficulty seeing how they got that answer.

Did I do something incorrectly? Or am I simplifying wrong?

Thank you

$\displaystyle x^2 - x - e = 0$

a = 1, b = -1, c = -e. These are what you're meant to substitute into the quadratic formula.
• Sep 3rd 2009, 04:10 AM
HallsofIvy
Quote:

Originally Posted by sleels
Hello, I need some help with this problem.

The problem is:

$\displaystyle Ln(x) + Ln(x-1) = 1$

Heres what I got
----------------

$\displaystyle Ln(x) + Ln(x-1) = 1]$

$\displaystyle Ln(x^2 - x) = 1$

$\displaystyle x^2 - x = e^1$

$\displaystyle x^2 - x - e = 0$

Now I use the quadratic equation to get

$\displaystyle \frac{x \pm \sqrt{x^2 - 4(x^2)(-e)}}{2x^2}$

The answer is supposed to be :
$\displaystyle \frac{(1 \pm \sqrt{1 + 4e})}{2}$

I'm having some difficulty seeing how they got that answer.

Did I do something incorrectly? Or am I simplifying wrong?

Thank you

You don't put the "x" inside the solution! The "a", "b", and "c" in the quadratic formula, $\displaystyle \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ are the numbers multiplying x and $\displaystyle x^2$. In $\displaystyle x^2- x- e= 0$, a= 1, b= -1, and c= -e.
• Sep 3rd 2009, 12:56 PM
sleels
doh! I knew it was something stupid.

Thanks a bunch