# Risolvi l’equazione di incognita $x$:

Risolvi l’equazione di incognita $x$:
$(p-1)x^2+px-2x=p$
assegnando a $p$ i valori:
a)$p=1/3$
b)$p=sqrt3$
c)$p=0,1$

Svolgimento
$(p-1)x^2+px-2x=p$;
$(p-1)x^2+(p-2)x-p=0$;

a)Per $p=1/3$ si ha:
$(1/3-1)x^2+(1/3-2)x-1/3=0$;
$-2/3x^2-5/3x-1/3=0$;
Moltiplicando ambo i membri per $3$ e cambiando di segno otteniamo
$2x^2+5x+1=0$

$\Delta=b^2-4ac=(5)^2-(4*1*(2))=25-8=17$
$x_(1,2)=(-b+-sqrt(\Delta))/(2a)=(-5+-sqrt(17))/4=> x_1=(-5-sqrt(17))/4 ^^ x_2=(-5+sqrt(17))/4$.

b)Per $p=sqrt3$ si ha:
$(sqrt3-1)x^2+(sqrt3-2)x-sqrt3=0$;

$\Delta=b^2-4ac=(sqrt3-2)^2-(4*sqrt3*(sqrt3-1))=3+4-4sqrt3+4sqrt3(sqrt3-1)=3+4-4sqrt3-4sqrt3+12=19-8sqrt3$
$x_(1,2)=(-b+-sqrt(\Delta))/(2a)=(-sqrt3+2+-sqrt(19-8sqrt3))/(2(sqrt3-1))=>$
$=> x_1=(-sqrt3+2+sqrt(19-8sqrt3))/(2(sqrt3-1)) ^^ x_2=(-sqrt3+2-sqrt(19-8sqrt3))/(2(sqrt3-1))$.

c)Per p=0,1=1/(10)
$(1/(10)-1)x^2+(1/(10)-2)x-1/(10)=0$;
$-9/(10)x^2-(19)/(10)x-1/(10)=0$;
Moltiplicando ambo i membri per $10$ e cambiando di segno otteniamo
$9x^2+19x+1=0$

$\Delta=b^2-4ac=(19)^2-(4*1*9)=361-36=325$
$x_(1,2)=(-b+-sqrt(\Delta))/(2a)=(-19+-sqrt(325))/(18)=> x_1=(-19+sqrt(325))/(18) ^^ x_2=(-19-sqrt(325))/(18)$.

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