Question

2x patrat - (3m+2)x +12=0 si 4xpatrat-(9m-2)x+36=0 au o radacina comuna Cum se rezolva?

Answer 1

2x^2-(3m+2)x+12=0
4x^2-(9m-2)x+36=0

x1+x2=3m+2
x1x2=12

x3+x4=9m-2
x3x4=36

Presupunem x2=x4

x1x2=12
x3x4=36, x2x3=36

(x1x2)/(x2x3)=12/36
x1/x3=1/3
x3=3x1

x1+x2=3m+2
x3+x4=9m-2, adica 3x1+x2=9m-2

x1+x2=3m+2
3x1+x2=9m-2
scadem prima din a doua:
3x1-x1=9m-2-3m-2
2x1=6m-4
x1=3m-2
x2=4
x3=9m-6
x4=4

2*16-(3m+2)*4+12=0
12m=36
m=3

4*16-(9m-2)*4+36=0
36m=108
m=3

Pt. m=3 ecuatiile au solutia x=4 comuna.

Answer 2

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