Eric Nicholas K's answer to Candice lim's Junior College 1 H2 Maths Singapore question.
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Good afternoon Candice! I found that the question's missing bracketing poses a challenge as to where the brackets could be.
I found that for y = x² + (3k - 2) (x + k) (x - 1), the graph can have no real root as is the case shown here.
This is because I obtained my discriminant of roots as
D = (3k - 2) (k - 1) times [(3k - 2) (k - 1) - 4k] and I found that there was no way I could combine them, and upon inspection, indeed I could not prove the "always real roots" case.
Instead, I tried a different bracketing arrangement y = x² + (3k - 2) x + k(k - 1) and found that on Desmos, the graph always has two real roots! Will work my answers based on this later.
I found that for y = x² + (3k - 2) (x + k) (x - 1), the graph can have no real root as is the case shown here.
This is because I obtained my discriminant of roots as
D = (3k - 2) (k - 1) times [(3k - 2) (k - 1) - 4k] and I found that there was no way I could combine them, and upon inspection, indeed I could not prove the "always real roots" case.
Instead, I tried a different bracketing arrangement y = x² + (3k - 2) x + k(k - 1) and found that on Desmos, the graph always has two real roots! Will work my answers based on this later.
Date Posted:
2 years ago
Thanks Mr Eric !
Totally agreed with your findings :)
Totally agreed with your findings :)
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