Question

x2+4x-5=0

Answer 1

\(x^2+4x-5=0\)

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

\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Answer 2

\(x^2+4x-5=0\)

\(\left(a=1;b=2;c=-5\right)\)

\(\Delta'=b^2-ac\)

\(\Delta'=2^2-1.\left(-5\right)\)

\(\Delta'=9>0\Rightarrow\sqrt{\Delta'}=\sqrt{9}=3\)

Vậy pt có 2 nghiệm phân biệt.

\(x_1=\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{\left(-2\right)+3}{1}=1\)

\(x_2=\dfrac{-b-\sqrt{\Delta'}}{a}=\dfrac{\left(-2\right)-3}{1}=-5\)

Vậy pt có tập nghiệm S = \(\left(1;-5\right)\)