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JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Determinants

The sum of the real roots of the equation

x 2 - 3 - 6 - 3 x 2 x - 1 x - 3 x + 2 = 0

is equal to

(A)-4
(B)0✓ Correct
(C)6
(D)1

Answer: (B)

Solution

Expanding the determinant and simplifying gives

- 5 x^{3} + 35 x - 30 = 0

, that is

x^{3} - 7 x + 6 = 0

. Factoring,

x^{3} - 7 x + 6 = (x - 1) (x - 2) (x + 3)

, so the roots are

x = 1, 2, - 3

, all of which are real. Their sum is

1 + 2 + (- 3) = 0

.

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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