Question

tìm x biết 

a) x² + x - 6 = 0

b) x² -2x - 3 =0 

 

Answer 1

a) \(x^2+x-6=0\)

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

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

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

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

b)\(x^2-2x-3=0\)

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

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

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

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

 

Answer 2

a) x² +x -6 = x² -2x +2x +x -6 = 0

x( x-2) + 3(x-2) = 0

(x-2)(x +3) = 0

x = 2 ; x = -3

b) x² -2x -3 = x² + x - x -2x -3 = 0

x(x-1) + 3(x-1) = 0

(x+1)(x-3) = 0

x = -1 ; x = 3

 

 

 

Answer 3

a) \(\Delta=1^2-4.1.\left(-6\right)=25>0\left(PTC2NPB\right)\)

\(X_1=\frac{-1+\sqrt{25}}{2}=2\)

\(X_2=\frac{-1-\sqrt{25}}{2}=-3\)

b) \(\Delta=\left(-2\right)^2-4.1.\left(-3\right)=16>0\left(PTC2NPB\right)\)

\(X_1=\frac{2+\sqrt{16}}{2}=3\)

\(X_2=\frac{2-\sqrt{16}}{2}=-1\)