Question

Trebuie rezolvate cu delta va rog mult puncte maxime și coronița
Nu va bateți joc!

Answer 1

1)

$\it x^2-3x+2=0,\ a=1,\ b=-3,\ c=2\\ \\ \Delta=b^2-4ac=(-3)^2-4\cdot1\cdot2=9-8=1\\ \\ \\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{3\pm1}{2} \Rightarrow\begin{cases}\ \it x_1=\dfrac{3-1}{2}=\dfrac{2}{2}=1\\ \\ \it x_2=\dfrac{3+1}{2}=\dfrac{4}{2}=2\end{cases}$

3)

$\it x^2-2x+1=0 \Rightarrow (x-1)^2=0 \Rightarrow x_1=x_2=1$

4)

$\it-x^2+x-7=0|_{\cdot(-1)} \Rightarrow x^2-x+7=0,\ a=1,\ b=-1,\ c=7\\ \\ \Delta =b^2-4ac=(-1)^2-4\cdot1\cdot7=1-28=-27<0 \Rightarrow x_{1,2}\not\in\mathbb{R}$

5)

$\it 2x^2-8x+1=0,\ a=2,\ b=-8,\ c=1\\ \\ \Delta=b^2-4ac=(-8)^2 -4\cdot2\cdot1=64-8=56\\ \\ x_{1,2}=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{8\pm\sqrt{56}}{2\cdot2}=\dfrac{8\pm\sqrt{4\cdot14}}{4}=\dfrac{8\pm2\sqrt{14}}{4}=\dfrac{2(4\pm\sqrt{14})^{(2}}{4}=\\ \\ \\ =\dfrac{4\pm\sqrt{14}}{2}\ \begin{cases} \it x_1=\dfrac{4-\sqrt{14}}{2}\\ \\ \\ \it x_2=\dfrac{4+\sqrt{14}}{2}\end{cases}$