To complete the square for the polynomial \( v^2 - 11v \), we follow these steps:
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Take the coefficient of the linear term (\(-11\)) and halve it: \[ -\frac{11}{2} = -5.5 \]
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Square this result: \[ (-5.5)^2 = 30.25 \]
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Now, add this value to the polynomial: \[ v^2 - 11v + 30.25 \]
So, the number that makes the polynomial a perfect-square quadratic is \( 30.25 \).
Putting it all together, the completed square is: \[ v^2 - 11v + 30.25 = \left(v - 5.5\right)^2 \]
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