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Which of the following reveals the minimum value for the equation 2x^2 − 4x − 2 = 0? 2(x − 1)^2 = 4 2(x − 1)^2 = −4 2(x − 2)^2 = 4 2(x − 2)^2 = −4

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we have the quadratic equation


2x^2-4x-2=0

Convert the given equation into vertex form

so

Factor 2


\begin{gathered} 2x^(2)-4x-2=0 \\ 2x^2-4x=2 \\ 2(x^2-2x)=2 \\ (x^2-2x)=1 \end{gathered}

Complete the square


\begin{gathered} (x^2-2x+1)=1+1 \\ rewrite\text{ as perfect square} \\ (x-1)^2=2 \\ 2(x-1)^2=4 \end{gathered}

The answer is the first option

Explanation factoring part


(x^2-2x)=1

Divide the term 2 by 2 and square it

2/2=1 -----> 1^2


\begin{gathered} (x^2-2x+1^2-1^2)=1 \\ (x^2-2x+1)=1+1 \\ (x^2-2x+1)=2 \end{gathered}

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