Question

Să se determine X€R astfel încât tripletele de numere sa fie în progresie aritmetica
a) x-1,x+5,2x+8
b)3x-7,5x-7,x²+2x-1
c) 2x²-9x-3,x²-3x-2,2x²-11x+19
d) 2x²-x-13,x²+x-4,2x²-3x+5

Answer 1

a) x+5=(x-1+2x+8)/2

2x+10=3x+7

10-7=3x-2x

3=x

b) 5x-7=(3x-7+x^2+2x-1)/2

10x-14=x^2+5x-8

x^2-5x+6=0

delta=25-24

=1

x1=5+1/2

=6/2

=3

x2=5-1/2

=4/2

=2

c) x^2-3x-2=(2x^2-9x-3+2x^2-11x+19)/2

2x^2-6x-4=4x^2-20x+16

2x^2-14x+20=0 :2

x^2-7x+10=0

delta=49-40

=9

x1= 7+3/2

=10/2

=5

x2=7-3/2

=4/2

=2

d) x^2+x-4=(2x^2-x-13+2x^2-3x+5)/2

2x^2+2x-8=4x^2-4x-8

2x^2-6x=0

2x(x-3)=0

x1=0

x2=3