<p>Number of real roots of the equation <span>\(\sum_{k=1}^{3} (x-k)^2 = 0\)</span> is</p>
Step-by-Step Solution
Key Concept: A sum of non-negative real numbers (squares) equals zero if and only if each term is individually zero. Since $(x-k)^2 \geq 0$ for all real $x$ and all $k$, the only way their sum can equal zero is if all three squared terms are zero simultaneously, which is impossible for a single value of $x$.
<p><strong>Step 1:</strong> Write out the equation explicitly.</p><p>$$\sum_{k=1}^{3} (x-k)^2 = 0$$</p><p>$$(x-1)^2 + (x-2)^2 + (x-3)^2 = 0$$</p><p><strong>Step 2:</strong> Recognize that each squared term is non-negative.</p><p>Since $(x-1)^2 \geq 0$, $(x-2)^2 \geq 0$, and $(x-3)^2 \geq 0$ for all real $x$, their sum is non-negative.</p><p><strong>Step 3:</strong> Apply the property that a sum of non-negative terms equals zero only when each term is zero.</p><p>For the sum to equal zero:</p><p>$(x-1)^2 = 0$ AND $(x-2)^2 = 0$ AND $(x-3)^2 = 0$</p><p><strong>Step 4:</strong> Analyze the contradiction.</p><p>These conditions would require:</p><p>$x = 1$ AND $x = 2$ AND $x = 3$</p><p>This is impossible for any single real value of $x$.</p><p><strong>Step 5:</strong> Conclude the number of real roots.</p><p>Since no real value of $x$ can satisfy all three conditions simultaneously, the equation has no real solutions.</p><p><strong>∴ Answer:</strong> P</p>
Correct Answer: P