Math Help - [SOLVED] Help on Quadratic Equation

1. [SOLVED] Help on Quadratic Equation

Hello, I need some help with this problem.

The problem is:

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

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

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

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

$x^2 - x = e^1$

$x^2 - x - e = 0$

Now I use the quadratic equation to get

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

The answer is supposed to be :
$
\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

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

The problem is:

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

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

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

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

$x^2 - x = e^1$

$x^2 - x - e = 0$

Now I use the quadratic equation to get

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

The answer is supposed to be :
$
\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
$x^2 - x - e = 0$

a = 1, b = -1, c = -e. These are what you're meant to substitute into the quadratic formula.

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

The problem is:

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

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

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

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

$x^2 - x = e^1$

$x^2 - x - e = 0$

Now I use the quadratic equation to get

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

The answer is supposed to be :
$
\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, $\frac{-b\pm\sqrt{b^2- 4ac}}{2a}$ are the numbers multiplying x and $x^2$. In $x^2- x- e= 0$, a= 1, b= -1, and c= -e.

4. doh! I knew it was something stupid.

Thanks a bunch