Question

giải phương trình:

x^2 +x = 4

Answer 1

\(x^2+x=4\)

\(\Leftrightarrow x^2+x-4=0\)

\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{17}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{17}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}\right)\left(x+\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}\right)=0\)

\(\Leftrightarrow x=\dfrac{-1-\sqrt{17}}{2}\) hay \(x=\dfrac{-1+\sqrt{17}}{2}\)

-Vậy \(S=\left\{\dfrac{-1-\sqrt{17}}{2};\dfrac{-1+\sqrt{17}}{2}\right\}\)