Explicație pas cu pas:
4.
(x+3)^2-25=x^2+6x+9-25
=x^2+6x-16
a=1
b=6
c=-16
Delta=b^2-4ac=36+64=100=>ecuația admite 2 soluții reale si distincte
Deci,
x1=(-b+radical delta) /2a=2
x2=(-b-radical detla) /2a=-8
Asadar, x^2+6x-16=(x-2)(x-8)
(x-4)(3x+7)-(x-4)(4x+2)+3(x+4)=3x^2+7x-12x-28-(4x^2+2x-16x-8)+3x+12=-x^2+12x+13
a=-1
b=+12
c=+13
Delta =196
X1= 1
X2=-13
(x-1)(x+13)
5. a). d=l radical 2=8 radical 2
b). Aplicam teorema lui Pitagora
AM^2+AD^2=DM^2
64+36=DM^2
DM=10 cm
c). A ABCD= A DCB+A DMB+ A DAM
A ABCD=8*8=64
A DCB=(8*8)2=32
A DAM=(8*6)/2=24
DECI
64=32+A DMB+24
A DMB=8
(2X+1)^2=4X^2+4X+1
