Question

Să se rezolve ecuațiile de gradul al doilea:x2+3x+2=0,x2-5x+6=0,x2-4x+4=0
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Answer 1

Răspuns:

Explicație pas cu pas:

$ax^{2} +bx+c=0\\\\delta=b^{2} -4ac\\\\x_{12} =\frac{-b+-\sqrt{delta} }{2a}$

x² + 3x + 2 = 0

a = 1, b = 3, c = 2

Δ = 3² - 4 · 1 · 2 = 9 - 8 = 1, √Δ = √1 = 1

$x_{1} =\frac{-3-1}{2} =\frac{-4}{2} =-2\\\\x_{2} =\frac{-3+1}{2} =\frac{-2}{2} =-1$

ș.a.m.d.

Answer 2

$\it x^2+3x+2=0,\ a=1,\ b=3,\ c=2\\ \\ \Delta=b^2-4ac=3^2-4\cdot2=9-8=1\\ \\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a} \Rightarrow x_{1,2}=\dfrac{-3\pm\sqrt1}{2}=\dfrac{-3\pm1}{2}\Rightarrow\begin{cases} \it x_1=\dfrac{-3-1}{2}=\dfrac{-4}{2}=-2\\ \\ \\ \it x_2=\dfrac{-3+1}{2}=\dfrac{-2}{2}=-1\end{cases}$

$\it Sau:\\ \\ x^2+3x+2=0 \Rightarrow x^2+2x+x+2=0 \Rightarrow x(x+2)+(x+2)=0 \Rightarrow \\ \\ \Rightarrow (x+2)(x+1)=0\Rightarrow \begin{cases}\it x+2=0 \Rightarrow x_1=-2\\ \\ \it x+1=0 \Rightarrow x_2=-1\end{cases}$

c)

$\it x^2-4x+4=0 \Rightarrow x^2-4x+2^2=0 \Rightarrow (x-2)^2=0 \Rightarrow x=2$