How many real roots does the equation \sqrt{x+9}=x-3 have?

Explanation

Squaring both sides gives x+9 = x^2-6x+9, which simplifies to x^2-7x=0. The roots are x=0 and x=7. Checking the original equation, x=0 gives 3=-3 (invalid), while x=7 gives 4=4. Thus, there is exactly one valid real root.

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