Show that the expression $\frac{(ax-b)(dx-c)}{(bx-a)(cx-d)}\\$ will be capable of all values when x is real, if $a^2-b^2$ and $c^2-d^2$ have the same sign.
Here's my approach:
I tried equating it with y which formed another Quadratic Equation, then after computing the Discriminant, here's what I got. If x is real, we must have $(ac+bd)^2(1-y)^2-4(ad-bcy)(bc-ady)$ positive I am stuck after that. Though I have the solution but it doesn't seem satisfactory.
A detailed answer would be helpful.
For reference this question is from Hall and Knight Higher Algebra, Ch-9, Examples 9B Q14.
This is the solution from the Solution's book of the treatise you quoted.
Well
we have that $$ \begin{gathered} \frac{{\left( {ax - b} \right)\left( {dx - c} \right)}} {{\left( {bx - a} \right)\left( {bx - a} \right)}} = y \hfill \\ \hfill \\ \left( {ad - bcy} \right)x^2 - \left( {ac + bd} \right)\left( {1 - y} \right)x + \left( {bc - ady} \right) = 0 \hfill \\ \end{gathered} $$ thus, in order to obtain a solution in $x$ it must be $\Delta \geq 0$. This means that $$ \left( {ac + bd} \right)^2 \left( {1 - y} \right)^2 - 4\left( {ad - bcy} \right)\left( {bc - ady} \right) \geqslant 0 $$ thus $$ \left( {ac + bd} \right)^2 \left( {1 - 2y + y^2 } \right) - 4\left( {adbc - a^2 d^2 y - b^2 c^2 y + adbcy^2 } \right) \geqslant 0 $$ thus $$ \left( {ac + bd} \right)^2 \left( {1 + y^2 } \right) - 2y\left( {ac + bd} \right)^2 - 4\left[ {adbc\left( {1 + y^2 } \right) - \left( {a^2 d^2 + b^2 c^2 } \right)y} \right] \geqslant 0 $$ thus $$ \left( {ac + bd} \right)^2 \left( {1 + y^2 } \right) - 2y\left( {ac + bd} \right)^2 - 4adbc\left( {1 + y^2 } \right) + 4\left( {a^2 d^2 + b^2 c^2 } \right)y \geqslant 0 $$ so that $$ \left[ {\left( {ac + bd} \right)^2 - 4adbc} \right]\left( {1 + y^2 } \right) - 2y\left[ {\left( {ac + bd} \right)^2 - 2\left( {a^2 d^2 + b^2 c^2 } \right)} \right] \geqslant 0 $$ Therefore we have $$ \begin{gathered} \left[ {a^2 c^2 + b^2 d^2 + 2adbc - 4adbc} \right]\left( {1 + y^2 } \right) + \hfill \\ - 2y\left[ {a^2 c^2 + 2abcd + b^2 d^2 - 2\left( {a^2 d^2 + b^2 c^2 } \right)} \right] = \hfill \\ \hfill \\ \left[ {a^2 c^2 + b^2 d^2 - 2adbc} \right]\left( {1 + y^2 } \right) + \hfill \\ - 2y\left[ {a^2 c^2 + 2abcd + b^2 d^2 - 4abcd + 4abcd - 2\left( {a^2 d^2 + b^2 c^2 } \right)} \right] = \hfill \\ \hfill \\ = \left[ {\left( {ac - bd} \right)^2 } \right]\left( {1 + y^2 } \right) - 2y\left[ {a^2 c^2 - 2abcd + b^2 d^2 - 2\left( {a^2 d^2 - 2abcdb^2 + b^2 c^2 } \right)} \right] = \hfill \\ \hfill \\ = \left[ {\left( {ac - bd} \right)^2 } \right]\left( {1 + y^2 } \right) - 2y\left[ {\left( {ac - bd} \right)^2 - 2\left( {ad - bc} \right)^2 } \right] = \hfill \\ \hfill \\ = \left( {ac - bd} \right)^2 y^2 - 2y\left[ {\left( {ac - bd} \right)^2 - 2\left( {ad - bc} \right)^2 } \right] + \left( {ac - bd} \right)^2 \geq 0 \hfill \\ \hfill \\ \end{gathered} $$ This is true for each $y$ if and only if $ \Delta _{\text{1}} = \left[ {\left( {ac - bd} \right)^2 - 2\left( {ad - bc} \right)^2 } \right]^2 - \left( {ac - bd} \right)^4 \leqslant 0 $ thus $$ \begin{gathered} \left( {ac - bd} \right)^4 - 4\left( {ac - bd} \right)^2 \left( {ad - bc} \right)^2 + 4\left( {ad - bc} \right)^4 - \left( {ac - bd} \right)^4 \leqslant 0 \hfill \\ \hfill \\ \end{gathered} $$ thus $$ - 4\left( {ac - bd} \right)^2 \left( {ad - bc} \right)^2 + 4\left( {ad - bc} \right)^4 \leqslant 0 $$ thus $$ \left( {ad - bc} \right)^2 \left[ { - \left( {ac - bd} \right)^2 + \left( {ad - bc} \right)^2 } \right] \leqslant 0 $$ thus $$ - \left( {ac - bd} \right)^2 + \left( {ad - bc} \right)^2 \leqslant 0 $$ thus $$ \left( {ac - bd} \right)^2 \geqslant \left( {ad - bc} \right)^2 $$ thus $$ \left[ {\left( {ac - bd} \right) - \left( {ad - bc} \right)} \right]\left[ {\left( {ac - bd} \right) + \left( {ad - bc} \right)} \right] \geqslant 0 $$ thus $$ \left[ {\left( {a + b} \right)\left( {c - d} \right)} \right]\left[ {\left( {a - b} \right)\left( {c + d} \right)} \right] \geqslant 0 $$ and finally $$ \left( {a^2 - b^2 } \right)\left( {c^2 - d^2 } \right) \geqslant 0 $$ This means that either the two factors have the same sign or at least one of them is zero. However, in this case, you have some degenerate situation.