Question

i ) \(x^2+2x-8=0\)

Answer 1

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

\(\left(x^2+2.x.1+1^2\right)-9=0\)

\(\left(x+1\right)^2=9\)

\(\Leftrightarrow\left(x+1\right)=\pm\sqrt{9}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=3\\x+1=-3\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-4\end{array}\right.\)

Answer 2

Theo định lí Viet ta có :

\(x_1+x_2=x\)

\(\Leftrightarrow x_1+x_2=-\frac{b}{a}\)

\(\Rightarrow x_1+x_2=-x\)

\(\Rightarrow x_1+x_2=-2\)

\(\Rightarrow x=2\)

\(x_1.x_2=\frac{x}{2}\)

\(x_1.x_2=\frac{c}{a}\)

\(x_1.x_2=-8\)

\(\Rightarrow x=-\frac{8}{2}=-4\)