Question

2x+n+2<=nx+5, n e R
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Answer 1

2x+n+2 <= nx+5 ,unde n∈R <=> nx+5-2x-n-2 >= 0 <=> nx-2x-n+3 >= 0 .

nx-2x-n+3 >= 0 <=> -2x+nx-n+3 >= 0 si observam ca membrul stang este de forma unei functii de gradul al II-lea ,unde f:R→R ,a.i. f(x)=ax²+bx+c ,unde a≠0 ,iar in cazul nostru avem

Δ=b²-4ac=(-2)²-4·n·(-n+3)=4-4·n·(-n+3)=4+4n²-12n=4n²-12n+4 > 0 <=>n²-3n+1 > 0 <=>

Δ=b²-4ac=(-3)²-4·1·1=9-4=5 > 0 ,care este adevarat <=> n₁=3-√5 /2 si n₂=3+√5 /2 =>

daca n∈(-∞,3 -√5 /2) ∪ (3 +√5 /2) .