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Functions help

I have done part 3 of this question, but I am stuck on the final part when I find the two values for k, K <0 and K < 64/9

Why is it that the answer is 0 < K < 64/9?

I thought we would draw the graph, and do K < 0, and K > 64/9, since the values must be outside of the graph?

Am I making any sense?

Anyways, I attached the question and mark scheme below.
(edited 2 years ago)
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For no solutions you need b2−4ac<0b^2-4ac < 0 i.e. (−3k)2−4(k)(16)<0(-3k)^2 - 4(k)(16) < 0.

Expand this quadratic and sketch it. You should see that it is negative BETWEEN its roots.
(edited 2 years ago)
Oh ok, I see what you mean. Thanks a lot!

Do you mind @RDKGames explaining how you wrote the b^2 - 4ac like that?
Original post by sweetescobar
Oh ok, I see what you mean. Thanks a lot!

Do you mind @RDKGames explaining how you wrote the b^2 - 4ac like that?

https://www.thestudentroom.co.uk/help/latex

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