Question

If all the solutions of the inequality $x^2-6 a x+5 a^2 \leq 0$ are also the solutions of inequality $x^2-14 x+40 \leq 0$ then find the number of possible integral values of a.

Answer 1

Let $f(x)=x^2-6 a x+5 a^2=(x-a)(x-5 a)$
and $g(x)=x^2-14 x+40=(x-4)(x-10)$
Clearly, $g(a) \leq 0$ and $g(5 a) \leq 0$ must satisfy simultaneously.
Now, $g(a) \leq 0 \Rightarrow a^2-14 a+40 \leq 0 \Rightarrow 4 \leq a \leq 10$...(1)

Also, $g (5 a ) \leq 0 \Rightarrow 25 a ^2-70 a +40 \leq 0$
$ \Rightarrow(5 a -4)( a -2) \leq 0 \Rightarrow \frac{4}{5} \leq a \leq 2$ ....(2)
$\therefore (1) \cap(2)$ gives $a \in \phi$.
Hence no integral value of a can satisfy it.