<p>Find the number of real roots of the equation \[\begin{vmatrix} 0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0 \end{vmatrix} = 0,\] where \(a \neq b \neq c\) and \(b(a+c) > ac\).</p>
Step-by-Step Solution
Key Concept: Recognize that the coefficient matrix is skew-symmetric (A^T = -A), so its determinant equals zero for all x when the matrix dimension is odd, but we must carefully analyze the actual polynomial structure that emerges from the expansion.
<p><strong>Step 1:</strong> Recognize the matrix is skew-symmetric: if we denote M(x), then the entries satisfy m_ij = -m_ji. The diagonal is identically zero.</p><p><strong>Step 2:</strong> Expand the determinant along the first row:</p><p>det(M) = 0·(minor) - (x-a)·∣x+a, x-c; x+b, 0∣ + (x-b)·∣x+a, 0; x+b, x+c∣</p><p><strong>Step 3:</strong> Calculate the 2×2 minors:</p><p>First minor: (x+a)·0 - (x-c)(x+b) = -(x-c)(x+b)</p><p>Second minor: (x+a)(x+c) - 0·(x+b) = (x+a)(x+c)</p><p><strong>Step 4:</strong> Substitute back:</p><p>det(M) = -(x-a)·[-(x-c)(x+b)] + (x-b)·[(x+a)(x+c)]</p><p>= (x-a)(x-c)(x+b) + (x-b)(x+a)(x+c)</p><p><strong>Step 5:</strong> Expand both products:</p><p>(x-a)(x-c)(x+b) = x³ + (b-a-c)x² + (-ac-ab+bc)x + abc</p><p>(x-b)(x+a)(x+c) = x³ + (a+c-b)x² + (ac-ab-bc)x - abc</p><p><strong>Step 6:</strong> Add them:</p><p>det(M) = 2x³ + (-ab-bc+ac-ab-bc+ac)x + 0</p><p>= 2x³ + 2(ac - ab - bc)x = 2x[x² + (ac - ab - bc)]</p><p><strong>Step 7:</strong> For roots: either x = 0 or x² = ab + bc - ac = b(a+c) - ac</p><p>Given condition: b(a+c) > ac, so b(a+c) - ac > 0</p><p>Therefore x² = positive value gives two more real roots: x = ±√[b(a+c) - ac]</p><p>∴ The equation has <strong>3 real roots</strong></p>
Correct Answer: C