First slide

Theory of expressions

Question

If roots of $x^{2} - (a - 3) x + a = 0$ are such that at least one of them is greater than 2, then

Moderate

A

$a \in [7, 9]$
B

$a \in [7, \infty)$
C

$a \in [9, \infty)$
D

$a \in [7, 9)$

Solution

$\begin{aligned} x^{2} - (a - 3) x + a = 0 \\ \Rightarrow D = (a - 3)^{2} - 4 a \\ = a^{2} - 10 a + 9 \\ = (a - 1) (a - 9) \end{aligned}$
Case I:
Both the roots are greater than 2.
$\begin{array}{l} D \geq 0, f (2) > 0, - \frac{B}{2 A} > 2 \\ \Rightarrow (a - 1) (a - 9) \geq 0; 4 - (a - 3) 2 + a > 0; \frac{a - 3}{2} > 2 \\ \Rightarrow a \in (- \infty, 1] \cup [9, \infty); a < 10; a > 7 \\ \Rightarrow a \in [9, 10) (1) \end{array}$
Case II:
One root is greater than 2 and the other root is less than or. equal to 2. Hence,
$\begin{array}{l} f (2) \leq 0 \\ \Rightarrow 4 - (a - 3) 2 + a \leq 0 \\ \Rightarrow a \geq 10 (2) \end{array}$
From (1) and (2),
$\begin{array}{l} a \in [9, 10) \cup [10, \infty) \\ \Rightarrow a \in [9, \infty) \end{array}$

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