Question

Exercitiul 11 .
Va rog!

Answer 1

2x²-3x+m+2 = 0

a) x1+x2+x1x2 < 0,5

3/2 + (m+2)/2 < 0,5 |•2

3+m+2 < 1

m < -4

b) x1²+x2² < x1x2+1

x1²+x2²-x1x2 < 1 |+3x1x2

x1²+x2²+2x1x2 < 1+3x1x2

(x1+x2)² < 1+3x1x2

9/4 < 1+3•(m+2)/2 |•4

9 < 4+6m+12

6m > -7

=> m < -7/6

c) (x1-1)(x2-1) < 0,25

f(x) = 2x²-3x+m+2 = (x-x1)(x-x2) = (x1-x)(x2-x)

f(1) = (x1-1)(x2-1) = 2-3+m+2 = m+1

=> m+1 < 0,25 => 4m+4 < 1 => 4m < -3 =>

=> m < -3/4