Question

Rezolvati in multimea numerelor reale ecuatia: x=radical din 4x+6

Answer 1

[tex]x= \sqrt{4x+6}\\\\ x^2=4x+6\\\\ x^2-4x-6=0\\\\ a=1\\ b=-4\\ c=-6\\\\ \·=b^2-4*a*c\\ \·=(-4)^2-4*1*(-6)\\ \·=16+24\\ \·=40\\ \\ x_1= \frac{-b- \sqrt{\·} }{2a}\\\\ x_1= \frac{-(-4)- \sqrt{40} }{2*1} = \frac{4-2 \sqrt{10} }{2} = \frac{2*(2-\sqrt{10})}{2}~=\bold{ 2- \sqrt{10} }\\\\\\ x_2= \frac{-b+ \sqrt{\·} }{2a}\\\\ x_2= \frac{-(-4)+ \sqrt{40} }{2*1} = \frac{4+2 \sqrt{10} }{2} = \frac{2*(2+ \sqrt{10})}{2}~=\bold{ 2+ \sqrt{10} }\\\\\\[/tex]

x>0 - conditie de existenta
deci, luam valoarea 2+√10