1. ## Properties of tangents

solve for k:

-x+k= 1/(x-1)

2. Originally Posted by math123456
solve for k:

-x+k= 1/(x-1)

$k=\frac{1}{x-1}+x$

I'm not sure if there's something elese you are supposed to do................

3. ## properties of tangents

i mean like solve for k using the quadratic formula.

4. Originally Posted by math123456
solve for k:

-x+k= 1/(x-1)

$
k = \frac{1}{x-1}+x
$

$
k = \frac{1+x(x-1)}{x-1}
$

$
k = \frac{x^2-x+1}{x-1}
$

5. I'm sorry math, but I'm not seeing how the quadratic formula has anthing to do with solving for k. Now, if we were solving for x, then the quadratic formula would be most appropriate.

K is not quadratic at all. In fact it is linear (or even more probable; a constant).
But for x...

$(x-1)(-x+k)=\frac{1}{x-1}*(x-1)$

$-x^2+kx+x-k=1$

$-x^2+kx+x-k-1=0$

$x^2-kx-x+k+1=0$

$x^2-(k+1)x+(k+1)=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Let a=1, b=-(k+1), and c=(k+1)

You can do the rest...

6. Originally Posted by VonNemo19
I'm sorry math, but I'm not seeing how the quadratic formula has anthing to do with solving for k. Now, if we were solving for x then the quadratic formula would be most appropriate,

K is not quadratic at all. In fact it is linear.
But for x...

$(x-1)(-x+k)=\frac{1}{x-1}*(x-1)$

$-x^2+kx+x-k=1$

$-x^2+kx+x-k-1=0$

$x^2-kx-x+k+1=0$

$x^2-(k-1)x+(k+1)=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Let a=1, b=-(k-1), and c=(k+1)

You can do the rest...
sign mistake in last step

$x^2-(k+1)x+(k+1)=0$

Let a=1, b=-(k+1), and c=(k+1)

7. Originally Posted by Shyam
sign mistake in last step

$x^2-(k+1)x+(k+1)=0$

Let a=1, b=-(k+1), and c=(k+1)

Right you are. Thank you.

8. Originally Posted by math123456
solve for k:

-x+k= 1/(x-1)
Is this the whole question, exactly as it's written in your book (or wherever it came from)? I doubt it very much since I'll bet diamonds to doughnuts that the correct answer is k = -1, 3. Next time please post the whole question so that people don't waste their time.