First slide

Theory of expressions

Question

If 6 lies between the roots of the equation $x^{2} + 2 (a - 3) x + 9 = 0$ then

Moderate

A

$a \in [- 3 / 4, \infty)$
B

$a \in (- \infty, - 3 / 4)$
C

$a \in (- \infty, 0) \cup (6, \infty)$
D

$a \in (- 3 / 4, 6)$

Solution

Let $f (x) = x^{2} + 2 (a - 3) x + 9$
If 6 lies between the roots of $f (x) = O,$ then we must have
Disc> 0, and (ii) f (6) < 0
$\begin{array}{l} \Rightarrow 4 (a - 3)^{2} - 36 > 0 \\ \Rightarrow (a - 3)^{2} - 9 > 0 \\ \Rightarrow a^{2} - 6 a > 0 \\ \Rightarrow a (a - 6) > 0 \Rightarrow a < 0 or a > 6 \\ and, f (6) < 0 \end{array}$
$\Rightarrow 36 + 12 (a - 3) + 9 < 0 \Rightarrow a < - \frac{3}{4}$
$a < - 3 / 4 i.e. a \in (- \infty, - 3 / 4)$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App