The number of real solutions to the equation $\sqrt{3x^2 - 18x + 52} + \sqrt{2x^2 - 12x + 162} = \sqrt{-x^2 + 6x + 280}$ is(are):
Step-by-Step Solution
Key Concept: Completing the square and substitution reduce the equation to a biquadratic form that must be verified in the original equation to eliminate extraneous solutions from squaring.
First, rewrite the expressions under the radicals by completing the square: $\sqrt{3(x-3)^2 + 25} + \sqrt{2(x-3)^2 + 144} = \sqrt{-(x-3)^2 + 289}$. Let $u = x-3$. The equation becomes $\sqrt{3u^2 + 25} + \sqrt{2u^2 + 144} = \sqrt{-u^2 + 289}$. For the RHS to be real, we need $u^2 \leq 289$, so $-17 \leq u \leq 17$. Squaring both sides and simplifying leads to $2\sqrt{(3u^2+25)(2u^2+144)} = -4u^2 + 280$. This requires $u^2 \leq 70$. Squaring again and reducing yields $u^4 - 70u^2 + 1369 = 0$, which factors as $(u^2 - 49)(u^2 - 28) = 0 + 49 \cdot 28 - 1369 = 0$. The solutions are $u = \pm 7$ and $u = \pm\sqrt{28}$. Verification shows only $u = 7$ and $u = -7$ satisfy the original equation, giving $x = 10$ and $x = -4$.
Correct Answer: 2