+1348 Let $f(x) = \frac{\sqrt{2x-6}}{x-3}$. Find the smallest integer value $a$ such that $f(a) = 1$.
+121 To find the smallest integer value \(a\) such that \(f(a) = 1\), we need to solve for \(a\) in the equation \(f(a) = 1\), where
\[f(x) = \frac{\sqrt{2x-6}}{x-3}\]
Substitute \(f(a) = 1\):
\[\frac{\sqrt{2a-6}}{a-3} = 1\]
Now, let's solve for \(a\):
Multiply both sides by \(a-3\):
\[\sqrt{2a-6} = a-3\]
Square both sides:
\[2a-6 = (a-3)^2\]
Expand the right side:
\[2a-6 = a^2 - 6a + 9\]
Move all terms to one side:
\[a^2 - 8a + 15 = 0\]
Factor the quadratic equation:
\[(a-5)(a+3) = 0\]
This gives us two possible values for \(a\): \(a = 5\) or \(a = -3\).
However, we need to find the smallest integer value of \(a\), so the answer is \(a = -3\).
Therefore, the smallest integer value \(a\) such that \(f(a) = 1\) is \(a = -3\).
