First slide

Theory of expressions

Question

If $a \in R$ and both the roots of $x^{2} - 6 a x + 9 a^{2} + 2 a - 2 = 0$ exceed 3, then a lies in the interval

Moderate

A

$(1, \infty)$
B

$(2, \infty)$
C

$(11 / 9, \infty)$
D

$ϕ$

Solution

Given equation can be written as $(x - 3 a)^{2} = 2 (1 - a)$
We must have $1 - a \geq 0 or a \leq 1$
Also, $x = 3 a \pm \sqrt{2} \sqrt{1 - a} .$
Both the roots will exceed 3, if smaller of the two roots viz
$3 a - \sqrt{2} \sqrt{1 - a}$ exceeds 3, that is, $3 a - \sqrt{2} \sqrt{1 - a} > 3$
$\Rightarrow 3 (a - 1) > \sqrt{2} \sqrt{1 - a}$ (1)

For $a = 1$ this is not possible. For $a < 1$ we can write (1)
as
$- 3 (\sqrt{1 - a})^{2} > \sqrt{2} \sqrt{1 - a}$
which is no possible
Thus, there is not real value of a.

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