Answer:
The correct option is: Both -5 and 1 are true solutions
Step-by-step explanation:
The given expression is:
√c^2+2c-4=√1-2c
By squaring both sides we get:
(√c^2+2c-4)^2=(√1-2c)^2
c^2+2c-4=1-2c
Move all the values to the L.H.S
c^2+2c-4-1+2c=0
c^2+4c-5=0
Now we will perform factorization:
c^2+5c-c-5=0
c(c+5)-1
(c+5)=0
(c+5)(c-1)=0
c+5 = 0 , c-1=0
c= 0-5 , c=0+1
c= -5 , c=1
Thus the correct option is Both -5 and 1 are true solutions....
2y+6x=0
-6x from both sides
2y=-6x+0
then divide by 2 to each side so it looks like this 2y/2=-6x+0/2
the 2 cancels 2y so you're left with y
Then divide -6/2=-3 and 0/2=0
y=-3x+0
Does that answer your question?
C.) I believe is the correct answer
Answer:
(x+6)^2+5
Step-by-step explanation:
