Multiple choice

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

  1. $6$
  2. $1$
  3. $0$
  4. $-4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding the determinant: x(-3x(x+2) - 2x(x-3)) + 6(2(x+2) - (-3)(x-3)) - 1(2(2x) - (-3)(-3x)) = 0. This simplifies to a cubic equation in x. The sum of roots of a polynomial ax^3 + bx^2 + cx + d = 0 is -b/a. Calculating the coefficient of x^2 and x^3 shows the sum is 0.

AI explanation

We evaluate the given determinant by expanding along the first row to get x((-3x)(x+2) - (x-3)(2x)) + 6(2(x+2) - (x-3)(-3)) - 1(2(2x) - (-3x)(-3)). Simplifying the expression inside the brackets yields x(-3x^2 - 6x - 2x^2 + 6x) + 6(2x + 4 + 3x - 9) - (4x - 9x), which further reduces to x(-5x^2) + 6(5x - 5) + 5x. Combining like terms results in the cubic equation -5x^3 + 35x - 30 = 0, which can be divided by -5 to get x^3 - 7x + 6 = 0. Factoring this polynomial gives (x - 1)(x - 2)(x + 3) = 0, revealing the real roots as 1, 2, and -3, which sum to 0.

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