Value of k, for an intersection of two parabolas

Question

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.

How do I do this?

If I did $x^2 -6x+14=-x^2 -20x -k$, I don't think it would work.

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$$x^2 -6x+14=-x^2 -20x -k$$ $$2x^2 +14x+14 = -k\\ -2x^2-14x-14=k$$

Now what?

Thank you

Answer 1

Wghat you obtain is a quadratic equation $$2x^2+14x+14+k=0$$ It has only one solution when the discriminant is equal to zero, meaning that $$14^2-4*2*(14+k)=0\to-84=8k\to k=-21/2$$

Answer 2

$x^2 - 6x + 14 = -x^2-20x - k \Rightarrow 2x^2 + 14x + 14+k = 0\Rightarrow \triangle = 0 \Rightarrow 14^2 - 4(2)(14+k) = 0$. Can you solve this equation for $k$?

Answer 3

This means $$ 0 = x^2 + 7 x + 7 + k/2 = (x + 7/2)^2 + 7 + k/2 - (7/2)^2 \iff \\ (x + 7/2)^2 = 49/4 - 28/4 - (2k)/4 = (21 - 2k) / 4 \iff \\ x =\frac{-7 \pm \sqrt{21-2k}}{2} $$ We have one solution $x = -7/2$ if the root does not contribute, thus $0 = 21 - 2k \iff k = 21/2$.

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