Question

Folosind forumla pe jumate rezolvati ecuatia (x+1)^2+x^2-5(x-1)^2=2(x-4) VA ROG FRUMOS !!!​

Answer 1

Răspuns:

(x+1)²+x²-5(x-1)²=2(x-4)

x²+x²-5(x²-2x+1)=2x-8

2x²-5x²+10x-5=2x-8

-3x²+10x-5-2x+8=0

-3x²+8x+3=0║·(-1)

3x²-8x-3=0

Formula   pe   jumate

b`=b/2= -8/2=-4

Δ `=(b`)²-ac

x1/2=(-b`±√Δ`)/a

Δ`=(_4)²-·3·(-3)=

16+9=25

x1=(4-√25)/3=

(4-5)/3= -1/3

x2=(4+√25)/3=(4+5)/3=9/3

x2=3

Explicație pas cu pas: