The given equation is
$$\eqalign{ & \left( {x - a} \right)\left( {x - b} \right) - 1 = 0\,,b > a. \cr & {\text{or }}{x^2} - \left( {a + b} \right)x + ab - 1 = 0 \cr & {\text{Let }}\,f\left( x \right) = {x^2} - \left( {a + b} \right)x + ab - 1 \cr} $$
Since co-eff. of $${x^2}$$ i.e. 1 > 0,
∴ it represents upward parabola, intersecting $$x$$ - axis at two points. (co-responding to two real roots, $$D$$ being $$+ve$$  ). Also
$$f\left( a \right) = f\left( b \right) = - 1$$
⇒ curve is below $$x$$ - axis at $$a$$ and $$b$$
⇒ $$a$$ and $$b$$ both lie between the roots.
Thus the graph of given $${\text{e}}{{\text{q}}^n}$$ is as shown.

from graph it is clear that one root of the equation lies in $$\left( { - \infty ,a} \right)$$  and other in $$\left( {b,\infty } \right)$$.

Let, $$y = \frac{{{x^2} - bc}}{{2x - b - c}}$$
$$\eqalign{ & \Rightarrow {x^2} - 2yx + \left( {b + c} \right)y - bc = 0 \cr & \because x \in R,\,\,{\text{so }}4{y^2} - 4\left( {b + c} \right)y + 4bc \geqslant 0 \cr & \Rightarrow x \leqslant b{\text{ or }}x \geqslant c\,\,\,\,\,\left( {\because b < c} \right) \cr} $$

It is clear from the graph that there is only one point of intersection of the curves $$y = \left| x \right|\,\,{\text{and }}y = {\log _{\frac{1}{2}}}x.$$
So, there is only one real solution

Let the discriminant of the equation $$x^2 + px + q = 0$$    by $$D_1 ,$$ then $${D_1} = {p^2} - 4q$$   and the discriminant $$D_2$$ of the equation $${x^2} + rx + s = 0$$    is $$D_2 = r^2 - 4s$$
$$\eqalign{ & \therefore {D_1} + {D_2} = {p^2} + {r^2} - 4\left( {q + s} \right) = {p^2} + {r^2} - 2pr\left[ {{\text{from the given relation}}} \right] \cr & \therefore {D_1} + {D_2} = {\left( {p - r} \right)^2} \geqslant 0 \cr} $$
Clearly at least one of $$D_1$$ and $$D_2$$ must be non-negative consequently at least one of the equation has real roots.

Multiplying the second equation by $$\frac{c}{{{a^3}}},$$
we get, $$\frac{{{b^2}{c^2}}}{{{a^3}}}{x^2} - \frac{{{b^2}c}}{{{a^2}}}x + c = 0$$
$$\eqalign{ & \Rightarrow a{\left( {\frac{{bc}}{{{a^2}}}x} \right)^2} - b\left( {\frac{{bc}}{{{a^2}}}} \right)x + c = 0 \cr & \Rightarrow \frac{{bc}}{{{a^2}}}x = \alpha ,\beta \cr & \Rightarrow \left( {\alpha + \beta } \right)\alpha \beta x = \alpha ,\beta \cr & \Rightarrow x = \frac{1}{{\left( {\alpha + \beta } \right)\alpha }},\frac{1}{{\left( {\alpha + \beta } \right)\beta }} \cr} $$

$$\eqalign{ & \frac{{2a + 3b}}{2} \geqslant \sqrt {2a \cdot 3b} \cr & \Rightarrow \,\,\frac{{ab}}{2} \geqslant \sqrt {6ab} \cr & \Rightarrow \,\,\sqrt {ab} \geqslant 2\sqrt 6 \cr & \Rightarrow \,\,ab \geqslant 24. \cr} $$

Let $$\alpha + i\beta ,\alpha - i\beta $$    be the roots. Then $${\alpha ^2} + {\beta ^2} = \frac{r}{p} > 0.$$    So, $$p, r$$  are of the same sign. Also $$p + r > 0.$$   So, $$p, r$$  are both positive.
If $$q < 0, p - q + r > 0.$$
If $$q > 0,{\left( {p + r} \right)^2} - {\left( {p - r} \right)^2} = 4pr \geqslant {q^2}\left( {\because \,\,{\text{roots are non - real}}} \right).$$
$$\eqalign{ & \therefore \,\,{\left( {p + r} \right)^2} \geqslant {q^2} + {\left( {p - r} \right)^2} \geqslant {q^2} \cr & \therefore \,\,p + r > q. \cr} $$

If $$\alpha , \beta$$  be the roots then
$$\alpha + \beta = - \frac{b}{a},\alpha \beta = \frac{c}{a}$$
Now the roots of $$\ell x^2 + mx + n = 0$$    are $$\frac{1}{\alpha },\frac{1}{\beta }$$
$$\eqalign{ & \therefore \frac{1}{\alpha } + \frac{1}{\beta } = - \frac{m}{\ell }{\text{and}}\frac{1}{\alpha } \cdot \frac{1}{\beta } = \frac{n}{\ell } \cr & {\text{or, }}\frac{{\alpha + \beta }}{{\alpha \beta }} = - \frac{m}{\ell }{\text{and}}\frac{a}{c} = \frac{n}{\ell } \cr & {\text{or, }} - \frac{b}{c} = - \frac{m}{\ell }{\text{and}}\frac{a}{c} = \frac{n}{\ell } \cr & {\text{or, }}\frac{a}{n} = \frac{b}{m} = \frac{c}{\ell }. \cr & \therefore a:b:c = n:m:\ell \cr} $$

$$\left( {{\lambda ^2} + \lambda - 2} \right){x^2} + \left( {\lambda + 2} \right)x - 1 < 0\,\,{\text{for all }}x \in R\,\,{\text{if }}D < 0\,\,{\text{and }}a < 0,\,{\text{i}}{\text{.e}}{\text{., }}{\left( {\lambda + 2} \right)^2} + 4\left( {{\lambda ^2} + \lambda - 2} \right) < 0\,\,{\text{and }}{\lambda ^2} + \lambda - 2 < 0.$$

both roots are less than 5 then (i) Discriminant $$ \geqslant 0$$
$$\eqalign{ & \left( {{\text{ii}}} \right)\,\,p\left( 5 \right) > 0 \cr & \left( {{\text{iii}}} \right)\,\,\frac{{{\text{Sum of roots}}}}{2} < 5 \cr} $$

$$\eqalign{ & {\text{Hence }}\left( {\text{i}} \right)4{k^2} - 4\left( {{k^2} + k - 5} \right) \geqslant 0 \cr & 4{k^2} - 4{k^2} - 4k + 20 \geqslant 0 \cr & 4k \leqslant 20 \cr & \Rightarrow \,\,k \leqslant 5 \cr & \left( {{\text{ii}}} \right)\,\,f\left( 5 \right) > 0;25 - 10k + {k^2} + k - 5 > 0 \cr & {\text{or }}\,{k^2} - 9k + 20 > 0 \cr & {\text{or }}k\left( {k - 4} \right) - 5\left( {k - 4} \right) > 0 \cr & {\text{or }}\left( {k - 5} \right)\left( {k - 4} \right) > 0 \cr & \Rightarrow \,\,k \in \left( { - \infty ,4} \right) \cup \left( { - \infty ,5} \right) \cr & \left( {{\text{iii}}} \right)\,\,\frac{{{\text{Sum of roots}}}}{2} = - \frac{b}{{2a}} = \frac{{2k}}{2} < 5 \cr} $$
The interection of (i), (ii) & (iii) gives
$$k \in \left( { - \infty ,4} \right)$$