# Thread: Determine the Value of K.

1. ## Determine the Value of K.

"If $f(x) = x^2 - 6x + 14$ and $g(x) = -x^2 - 20x - k$, determine the value of $k$ so that there is exactly one point of intersection between the two parabolas."

This is what I've done so far:

$f(x) + g(x) = 0$
$-26x + 14 - k = 0$
$a = 0, b = -26, c = (14 - k)$
$b^2 - 4ac = 0$
$(-26)^2 - 4(0)(14 - k) = 0$
$676 - 0 = 0$

Based on the fact that this method cannot solve for $k$, I've come to believe that this attempt at a solution was wrong. Though I can't think of any other way of solving this. Help would be appreciated.

"If $f(x) = x^2 - 6x + 14$ and $g(x) = -x^2 - 20x - k$, determine the value of $k$ so that there is exactly one point of intersection between the two parabolas."

This is what I've done so far:

$f(x) + g(x) = 0$

Why plus?! The two parabolas meet when $f(x)=g(x)\Longleftrightarrow f(x)-g(x)=0$ ...!

Tonio

$-26x + 14 - k = 0$
$a = 0, b = -26, c = (14 - k)$
$b^2 - 4ac = 0$
$(-26)^2 - 4(0)(14 - k) = 0$
$676 - 0 = 0$

Based on the fact that this method cannot solve for $k$, I've come to believe that this attempt at a solution was wrong. Though I can't think of any other way of solving this. Help would be appreciated.
.

3. Thanks for the correction. Got a final answer of 10.5.