The least positive value of
Question:

The least positive value of ‘ $a$ ‘ for which the equation, $2 x^{2}+(a-10) x+\frac{33}{2}=2 a$ has real roots is

Solution:

Since, $2 x^{2}+(a-10) x+\frac{33}{2}=2 a$ has real roots,

$\therefore \quad D \geq 0$

$\Rightarrow \quad(a-10)^{2}-4(2)\left(\frac{33}{2}-2 a\right) \geq 0$

$\Rightarrow \quad(a-10)^{2}-4(33-4 a) \geq 0$

$\Rightarrow a^{2}-4 a-32 \geq 0$

$\Rightarrow \quad(a-8)(a+4) \geq 0$

$\Rightarrow \quad a \leq-4 \cup a \geq 8$

$\Rightarrow \quad a \in(-\infty,-4] \cup[8, \infty)$

Administrator

Leave a comment

Please enter comment.
Please enter your name.