First slide

Theory of expressions

Question

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

Difficult

A

$a \in [7, 9]$
B

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

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

$a \in [7, 9)$

Solution

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

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