Question

Sa se determine x∈ R astfel incat urmatoarele numere sa fie in progresie aritmetica:
a)x-1 , $\sqrt{9x+4}$ , 2x
b)x-2 , $\sqrt{5x+1}$ ,x+4

Answer 1

a) prog aritmetica=> $\sqrt{9x+4}$=$\frac{x-1+2x}{2} = \frac{3x-1}{2}$

$\sqrt{9x+4}= \frac{3x-1}{2} |^{2}$=> 9x+4=$\frac{(3x-1)^{2} }{4}$=>(9x+4)*4=9x²-6x+1
36x+16-9x²+6x-1=0
-9x²+42x+15=0 |:3
-3x²+14x+5=0
Δ=14²-4*(-3)*5=196+60=256
√Δ=16.
  $x1= \frac{-14+16}{-6}= \frac{2}{-6}= -\frac{1}{3}$
x2=$\frac{-14-16}{-6}= \frac{-30}{-6} =5$
b) $\sqrt{5x+1}= \frac{x-2+x+4}{2}= \frac{2x+2}{2}=> \sqrt{5x+1}= x+1 |^{2}$    am scris asa fiindca am simplificat prin 2 fractia 2x+2 supra 2 (se poate simplifica fiindca daca dai factor comun ai 2(x+1) totul supra 2)
5x+1=x²+2x+1=>x²+2x+1-5x-1=0
x²-3x=0
x(x-3)=0⇒ x=0
                 x-3=0⇒x=3