Question

Sa se arate ca oricare ar fi X apartine R, sunt adevarate inegalitatile :
a. x^2+4x+5+( mai > sau = ) 1

b. x^2+10x+27 (mai > sau = ) 2

c. 2x^2+4x+5 (mai > sau = ) 3

Answer 1

x²+4x+5≥1⇒
x²+4x+5-1≥0⇒
x²+4x+4≥0⇒
a=1
b=4
c=4
d= b²-4ac⇒d=16-4×1×4⇒d=16-16⇒d=0
x1x2=-b+-√d/2a
x1=-4+0/2⇒-4/2⇒x1=-2.
x2=-2

x²+10x+27≥2⇒
x²+10x+27-2≥0⇒x²+10x+25≥0

a=1
b=10
c=25
d=b²-4ac⇒d=100-100⇒d=0
x1.x2=-10-+0/2⇒x1.x2=-5

2x²+4x+5≥3⇒
2x²+4x+5-3≥0⇒2x²+4x+2≥0
a=2
b=4
c=2
d=b²-4ac⇒d=16-16⇒d=0
x1 x2=-b+-√d/2a⇒x1x2=-4+-0/4⇒x1 x2=-1