Question

The sum of the real roots of the equation \(\left| {\begin{array}{*{20}{c}} x&{ - 6}&{ - 1}\\ 2&{ - 3x}&{x - 3}\\ { - 3}&{2x}&{x + 2} \end{array}} \right| = 0\), is equal to:
1. 6
2. 0
3. 1
4. -4

Answer 1

Correct Answer - Option 2 : 0

\({\rm{Given\;}}\left| {\begin{array}{*{20}{c}} x&{ - 6}&{ - 1}\\ 2&{ - 3x}&{x - 3}\\ { - 3}&{2x}&{x + 2} \end{array}} \right| = 0\) 

On expanding,

x(-3x × (x + 2) - 2x(x - 3)) - (-6)(2(x + 2) + 3(x - 3)) + (-1)(4x + 3(-3x))

⇒ x(-3x² - 6x - 2x² + 6x) + 6(2x + 4 + 3x - 9) - (4 x - 9x) = 0

⇒ x(-5x²) + 6(5x - 5) - 4x + 9x = 0

⇒ -5x³ + 30x - 30 - 4x + 9x = 0

⇒ 5x³ - 30x + 30 + 4x - 9x = 0

⇒ 5x³ - 35x + 30 = 0

⇒ 5(x³ - 7x + 6) = 0

⇒ x³ - 7x + 6 = 0

∵ All the roots are real.

∴ Sum of real roots = 1 - 7 + 6 = 0