Question

\(2x^2+5x-3=0\)

Giải pt

Answer 1

x₁=\(\dfrac{1}{2}\)

x₂=-3

Answer 2

\(2x^2+5x-3=0\)

\(\Leftrightarrow2x^2-x+6x-3=0\)

\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Answer 3

\(2x^2+5x-3=0\\ \Leftrightarrow2x^2+6x-x-3=0\\ \Leftrightarrow2x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)