Question

Putin ajutor?Ecuatie de gradul al II lea!
A) 2x^2+9x+10=0
B)x^2-5x+4=0
C)3x^2+x√7-2=0

Answer 1

   
[tex]\displaystyle\\ \texttt{\bf Formula pentru rezolvarea ecuatiei de gradul 2 este:}\\\\ x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\ \texttt{unde:}\\ a=\text{coeficientul lui }x^2\\ b=\text{ coeficientul lui }x\\ c=\text{ termenul liber }\\\\ \texttt{\bf Rezolvare:} [/tex]


[tex]\displaystyle\\ A)2x^2+9x+10=0\\\\ x{12}=\frac{-9\pm\sqrt{9^2-4\cdot2\cdot10}}{2\cdot2}=\frac{-9\pm\sqrt{81-80}}{4}=\frac{-9\pm\sqrt{1}}{4}=\frac{-9\pm1}{4}\\\\ x_1=\frac{-9+1}{4}=\frac{-8}{4}=\boxed{\bf-2}\\\\ x_2=\frac{-9-1}{4}=\frac{-10}{4}=\boxed{-\frac{5}{2}}[/tex]


[tex]\displaystyle\\ B) x^2-5x+4=0\\\\ x_{12}=\frac{5\pm\sqrt{(-5)^2-4\cdot1\cdot 4}}{2\cdot1}=\frac{5\pm\sqrt{25-16}}{2}=\frac{5\pm\sqrt{9}}{2}=\frac{5\pm3}{2}\\\\ x_1=\frac{5+3}{2}=\frac{8}{2}=\boxed{4} x_2=\frac{5-3}{2}=\frac{2}{2}=\boxed{1} [/tex]


[tex]\displaystyle\\ C)~3x^2+x\sqrt{7}-2=0\\\\ x_{12}=\frac{-\sqrt{7}\pm\sqrt{\Big(\sqrt{7}\Big)^2-4\cdot 3 \cdot (-2)}}{2\cdot 3}= \frac{-\sqrt{7}\pm\sqrt{7+24}}{6}=\frac{-\sqrt{7}\pm\sqrt{31}}{6}\\\\ x_1 = \boxed{\frac{-\sqrt{7}+\sqrt{31}}{6}}\\\\ x_2 = \boxed{\frac{-\sqrt{7}-\sqrt{31}}{6}} [/tex]