The number of real roots of the equation $\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$, is:
Step-by-Step Solution
Key Concept: Factor the expressions under the radicals: $x^2-4x+3=(x-1)(x-3)$, $x^2-9=(x-3)(x+3)$, $4x^2-14x+6=2(x-3)(2x-1)$. Factor out $\sqrt{x-3}$.
$\sqrt{(x-3)}[\sqrt{x-1}+\sqrt{x+3}] = \sqrt{(x-3)}\cdot\sqrt{4(2x-1)}$. Case 1: $x=3$ (valid). Case 2: $\sqrt{x-1}+\sqrt{x+3}=\sqrt{4x-2}$, squaring and solving gives $x=7/6$ which is rejected (domain issue). Only 1 real root.
Correct Answer: 2