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How to show that expression $$\frac{px^{2}+3x-4}{p+3x-4x^2}$$ will be capable of all values when $x$ is real,provided that $p$ has any value between $1$ and $7$?

Regarding my personal attempts,they are all futile.

I tried to expand $y=(px^{2}+3x-4)/(p+3x-4x^2$) to get a relation between $y$ and $p$ using "$b^{2}-4ac\geq0$"formula and substitute y with $1$.But,all futile.

  • will be capable ?? – alkabary Jun 12 '15 at 09:01
  • Do you mean the expression $\frac{px^2+3x-4}{p+3x-4x^2}$? – Rory Daulton Jun 12 '15 at 09:02
  • If $p$ is not between $1$ and $7$, then the expression is not capable of attaining the value $-1$. – Arthur Jun 12 '15 at 09:09
  • @RoryDaulton,I mean the same.But being Ignorant of LaTex ,I am unable to edit. –  Jun 12 '15 at 09:11
  • @Arthur,How did you figure out that? –  Jun 12 '15 at 09:21
  • Set $\frac{px^2+3x-4}{p+3x-4x^2} = -1$, get $(p-4)x^2 + 6x + (p-4) = 0$. There is a solution to this equation iff $\sqrt{36 - 4(p-4)^2} = \sqrt{-4p^2 + 32p-28} = 2\sqrt{-p^2 + 8p - 7}$ exists. That square root exists when $-p^2 + 8p - 7$ is non-negative, which is exactly when $p$ is between $1$ and $7$ (inclusive). – Arthur Jun 12 '15 at 09:29
  • @Arthur You should have given it as an answer.Thanks for the help. –  Jun 12 '15 at 09:39
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    It's not a full answer (there is no proof that you can reach other values for the relevant $p$ values). Lab Bhattacharjee's answer below wants you to do the same, only not just for $-1$, but for all possible $y$ values at the same time. – Arthur Jun 12 '15 at 09:42

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Satrting from $$y=\frac{px^{2}+3x-4}{p+3x-4x^2}$$ rewrite the equation as a quadratic in $x$; this gives $$ (p+4 y)x^2-3 (y-1)x-(4+py)=0$$ Compute the discriminant $$\Delta_x=(16 p+9) y^2+\left(4 p^2+46\right) y+16 p+9$$ Compute again the discriminant for this quadratic in $y$ $$\Delta_y=p^4-41 p^2-72 p+112$$ and this one must be negative or zero.

By inspection and succesive divisions, you could notice that this factors $$\Delta_y=(p-7) (p-1) (p+4)^2$$

I am sure that you can take from here.

Claude Leibovici
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