Question

Rezolvati in multimea nr reale ecuatia :$(3x+2 )^{2} =4$

Answer 1

Folosesti formula : $( a+b)^{2} = a^{2}+ 2ab + b^{2}$
Si rezulta :  $(3x)^{2}+2*3x*2+ 2^{2} =4$
$9x^{2} + 12x + 4 = 4$
$9x^{2} + 12 = 0$
deci e ecutatie de gradul 2.
a = 9 ; b = 12 ; c = 0
discriminantul ecuatiei delta ( notez cu D )
$D= b^{2}-2ac => D= 12^{2}-2*9*0= 12^{2} = 144$
$x1 = \frac{-b+ \sqrt{D} }{2a} => \frac{-12+ \sqrt{144} }{18} = \frac{-12+12}{18} = \frac{0}{18} = 0$
$x2= \frac{-12- \sqrt{144} }{18} = \frac{-12-12}{18} = \frac{24}{18} = \frac{4}{3}$

sa-mi spui daca-i corect :o3