JEE PYQ: Quadratic Equation Question 28
Question 28 - 2020 (08 Jan Shift 1)
The least positive value of ‘$a$’ for which the equation, $2x^2 + (a - 10)x + \frac{33}{2} = 2a$ has real roots is _______.
Show Answer
Answer: (8)
Solution
Since, $2x^2 + (a-10)x + \frac{33}{2} - 2a$ has real roots,
$\therefore D \geq 0$
$\Rightarrow (a-10)^2 - 4(2)\left(\frac{33}{2} - 2a\right) \geq 0$
$\Rightarrow (a-10)^2 - 4(33 - 4a) \geq 0$
$\Rightarrow a^2 - 4a - 32 \geq 0$
$\Rightarrow (a - 8)(a + 4) \geq 0$
$\Rightarrow a \leq -4 \cup a \geq 8$
$\Rightarrow a \in (-\infty, -4] \cup [8, \infty)$
BETA
