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$(x+2)(3x-14)=1/3((29)/2x+35)+((2x+7)x)/3+5x(x/2+1)$ |
di Francesco Speciale
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$(x+2)(3x-14)=1/3((29)/2x+35)+((2x+7)x)/3+5x(x/2+1)$
$(x+2)(3x-14)=1/3((29)/2x+35)+((2x+7)x)/3+5x(x/2+1)$;
$3x^2-14x+6x-28=(29)/6x+(35)/3+(2x^2+7x)/3+5/2x^2+5x$;
Il m.c.m. è $6$
$(18x^2-84x+36x-168)/6=(29x+70+2(2x^2+7x)+15x^2+30x)/6$;
Moltiplicando ambo i membri per $6$ si ha:
$18x^2-84x+36x-168=29x+70+4x^2+14x+15x^2+30x$;
Semplificando
$(18-4-15)x^2+(36-84-29-14-30)x-168-70=0$;
$-x^2-121x-238=0$ cioè
$x^2+121x+238=0$
$\Delta=b^2-4ac=(121)^2-(4*238*1)=14641-952=13689$
$x_(1,2)=(-b+-sqrt(\Delta))/(2a)=(-121+-sqrt(13689))/2=(-121+-(117))/2 => x_1=-2 ^^ x_2=-119$.
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