Question

Please solve with explanation

Answer 1

k=0 or 4

Explanation:

`y=x^2+(k+4)x+3k+4\\=x^2+(k+4)x+((x+4)/(2))^2 +3k+4-((x+4)/(2))^2\\=(x+(x+4)/(2))^2+3k+4-((x+4)/(2))^2\\`

so the vertex is at

`(-(k + 4)/(2), 3k+4-((x+4)/(2))^2)`

The vertex lies in the x-axis, so

`3k+4-((k+4)/(2))^2 = 0\\4(3k+4)-(k+4)^2=0\\12k+16-k^2-8k-16=0\\-k^2+4k=0\\-k(k-4)=0\\k-0, 4`