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Let $f(x) = \frac{\sqrt{2x-6}}{x-3}$.  Find the smallest integer value $a$ such that $f(a) = 1$.

 Aug 17, 2023
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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 \(a = -3\).

 Aug 17, 2023

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