Question

Puncte + coroana repede.....................................

Answer 1

Răspuns:

$ax^{2}+bx+c=0 \\D=b^{2}-4ac\geq 0 \\x_{1} +x_{2} = \frac{-b-\sqrt{D} }{2a} + \frac{-b+\sqrt{D} }{2a}=-2b/2a=-b/a\\x_{1} x_{2} = \frac{-b-\sqrt{D} }{2a} \frac{-b+\sqrt{D} }{2a}=-\frac{D-b^{2} }{4a^{2} }=-\frac{b^{2}-4ac-b^{2} }{4a^{2} }=\frac{c}{a}$

1)

$x^{2}-(2m+1)x+3m=0 \\x_{1} +x_{2} =(2m+1)/1=2m+1\\x_{1} x_{2} =\frac{3m}{1} \\x_{1} +x_{2} +x_{1} x_{2} =11\\=>2m+1+3m=11 =>5m=10=>m=2$

2)

$x^{2}+mx+2=0 \\(x_{1} +x_{2})^{2} -2x_{1} x_{2} =5=>m^{2} -2X2=5=>m^{2} =1=>m =+/-1$

procedeaza la fel si pentru celelalte puncte