Question

RAIZ DAS EQUACOES 

(X+3)2=49

(5X-4)2=9

Answer 1

(X+3)2=49
2x + 6 = 49
2x = 49 - 6
2x = 43
x = 43/2

(5X-4)2=9
10x - 8 = 9
10x = 9 + 8
10x = 17
x=17/10

Answer 2

OLAÁ DENISE

(X+3)2=49 
x²+3X+3X+9=0
x²+6X+9=0

 utilizando Bhaskara, temos
a=1,b=6 e c=9
Δ = b² -4ac
Δ = (6)² -4 .1.  9
Δ = 36-36= 0


x1 E X 2= (-b +/-√∆)/ 2a
-6/2 = -3


S={ -3}

(5X-4)2=9
25X²-40X+16 
Δ= x²-4.A.C
Δ=1600-1600
Δ=0
x1e x2= 40/50=4/5


=S={ 4/5}