Reference to book: Tranter, C.J., 1971, Advanced Level Pure Mathematics. The English Universities Press. London. (20)
Difficulties on finding the real values for this equation:
Show that for all real values of y , the expression $$ \frac{3y^2 - 2y -1 }{y^2 + y + 2}$$ always lie between $-4/7$ and $4$
I have tried $3y^2 -2y -1 = y^2 + y + 2$ and then combining into quadratic to form $6y^2 - 3y - 3.$
How do I express this into bracketed form?
How to find $(y-\alpha)(y-\beta)$ when $ny$ is involved in the equation?
Background: Self-study, mainly quantitative skills in applied mathematics very little so in analysis or pure mathematics. equivalent to early A-level/mid.
$$4-\frac{3y^2 - 2y -1 }{y^2 + y + 2} =\frac{y^2+6y+9}{y^2 + y + 2}=\frac{(y+3)^2}{(y+1/2)^2+7/4}\ge0.$$ I'll leave you to figure out the other inequality.