Question

4x^2+8x+2=0

Answer 1

\(4x^2+8x+2=0\)

\(\Leftrightarrow2\left(2x^2+4x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)^2=0\)

\(\Leftrightarrow2x+1=0\)

\(\Leftrightarrow2x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)

Answer 2

<=>4x(x+2)=-2

<=>4x=0 hay x+2=0

<=>X=0 hay x=-2

Answer 3