Question

The sum of the real roots of the equation | x 6 1 |
| 2 3x (x - 3)| = 0 | 3 2x (x = 2)|
is equal to (A) -4 (B) 0
(C) 6 (D) 1

Answer 1

$\left|\begin{array}{ccc}x&6&1\\2&3x&x-3\\3&2x&x-2\end{array}\right|=0$

$\left|\begin{array}{ccc}x&6&1\\2&3x&x-3\\1&-x&-1\end{array}\right|=0$ $R_3\implies\,R_3-R_2$

$\text{Expanding along first row}$

$x(-3x+x(x-3))-6(-2-x+3)+1(-2x-3x)=0$

$x(-3x+x^2-3x)-6(-x+1)+1(-5x)=0$

$x(-6x+x^2)+6x-6-5x=0$

$-6x^2+x^3+x-6=0$

$x^2(x-6)+1(x-6)=0$

$(x^2+1)(x-6)=0$

$x^2+1=0\;\text{or}\;x-6=0$

$x^2+1=0\;\text{or}\;x=6$

$x^2+1=0\;\text{gives unreal roots}$

$\therefore\textbf{The only real root is 6}$

$\implies\textbf{The sum of the real roots is 6}$

$\textbf{Option (C) is correct}$

Answer 2

The sum of real roots of given equation is 6

option (C)

Step-by-step explanation:

$\left|\begin{array}{ccc}x&6&1\\2&3x&x-3\\1&-x&-1\end{array}\right|=0 R_3\implies\,R_3-R_2$

$\text{We can expand through any Row and Column}$

$\text{But we choose the Row or Column with minimum unknowns and maximum zeroes}$

$\text{So we will expand through Row I}$

$x(3x(-1) - (-x)(x - 3)) - 6(2(-1) - 1(x - 3)) + 1(2(-x) - 1(3x)) =0-6x^2+x^3+x-6=0$

$\text{On expanding each bracket and solving above equation, we get:-}$

$x^2(x-6)+1(x-6)=0$

$\textbf{Taking (x-6) common , we get :-}$

$(x^2+1)(x-6)=0$

$\text{Equation will become zero if}$

$x-6=0\;\text{or}\;x^2+1=0$ $\text{or}$

$x=6\;\text{or}\;x^2+1=0$

$x^2+1=0\;\text{It gives unreal roots because square root of real number can not be negative}$

$x^2+1=0\;\textbf{gives unreal roots}$

$\textbf{ So, x=6 will be the only real root}$

$\textbf{ So, the sum of real roots of this equation is 6}$

$\textbf{So correct option is C}$