Răspuns:
Explicație pas cu pas:
a)2x²-x+2=0⇒ Δ=1-4×4<0 ecuatia nu are solutii reale
b)x²-4x+4=(x-2)² ecuatia are solutie unica x=2 Δ=16-16=0
[tex]\it a)\ 2x^2-x+2=0|_{\cdot2} \Rightarrow 4x^2-2x+4=0 \Rightarrow 3x^2+x^2-2x+1+3=0 \Rightarrow \\ \\ \Rightarrow 3x^2+3+(x-1)^2=0\ \ \ \ \ (*)\\ \\ 3x^2+3>0,\ \ (x-1)^2\geq0 \stackrel{(*)}{\Longrightarrow}\ ecua\c{\it t}ie\ imposibil\breve{a} \Rightarrow S=\varnothing\\ \\ \\ b)\ x^2-4x+4=0 \Leftrightarrow(x-2)^2=0 \Leftrightarrow x=2[/tex]
