Question

If all the solution 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 value 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, $g(x) \leq 0 \Rightarrow a^{2}-14 a+40 \leq 0$
$\Rightarrow \quad 4 \leq a \leq 10$....(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)
$\therefore (1) \cap(2) \text { gives } a \in \phi$
Hence no integral value of a can satisfy it.