First slide

Introduction to Determinants

Question

The sum of the real roots of the equation $|\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = 0$ is equal to

Easy

A

0
B

-4
C

2
D

1

Solution

$W e h a v e |\begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix}| = 0$
$\Rightarrow \begin{array}{l} \Rightarrow x (- 3 x^{2} - 6 x - 2 x^{2} + 6 x) + 6 (2 x + 4 + 3 x - 9) - 1 (4 x - 9 x) = 0 \\ x^{3} - 7 x + 6 = 0 \\ T h e r o o t s a r e 1, 2, - 3 . \end{array}$
$∴ s u m o f r o o t s = 1 + 2 - 3 = 0$

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