<p>a, b, c are real numbers with \(a^2 + b^2 + c^2 > 0\). Then the equation \(x^2 + (a+b+c)x + (a^2+b^2+c^2) = 0\) has</p>
<p>(a) 2 negative real roots</p>
<p>(b) 2 positive real roots</p>
<p>(c) 2 real roots with opposite signs</p>
<p>(d) no real roots</p>
Step-by-Step Solution
Key Concept: For a quadratic to have real roots, its discriminant must be non-negative. Here, the discriminant is (a+b+c)² - 4(a²+b²+c²), which can be rewritten using the expansion (a+b+c)² = a²+b²+c²+2(ab+bc+ca), leading to -3(a²+b²+c²)+2(ab+bc+ca). By recognizing this as -[(a-b)²+(b-c)²+(c-a)²], we see it's always non-positive.
<p><strong>Step 1:</strong> For the quadratic x² + (a+b+c)x + (a²+b²+c²) = 0, compute the discriminant:</p><p>Δ = (a+b+c)² - 4(a²+b²+c²)</p><p><strong>Step 2:</strong> Expand (a+b+c)²:</p><p>Δ = a² + b² + c² + 2ab + 2bc + 2ca - 4a² - 4b² - 4c²</p><p>Δ = -3a² - 3b² - 3c² + 2ab + 2bc + 2ca</p><p><strong>Step 3:</strong> Rewrite by grouping:</p><p>Δ = -(a² - 2ab + b²) - (b² - 2bc + c²) - (c² - 2ca + a²)</p><p>Δ = -[(a-b)² + (b-c)² + (c-a)²]</p><p><strong>Step 4:</strong> Since squares are non-negative and a²+b²+c² > 0 (at least one of a,b,c is non-zero), we have (a-b)² + (b-c)² + (c-a)² ≥ 0. Therefore Δ ≤ 0, with equality only if a = b = c = 0 (contradicting the given condition).</p><p>Thus Δ < 0 always.</p><p>∴ Answer: The equation has <strong>no real roots</strong> (two complex conjugate roots).</p>
Correct Answer: D