What is the solution to the system A. (-3,1) B. (0,3 C. (1,3) D. (1,-1)
2 answers:
You might be interested in
Answer:
Step-by-step explanation:
They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.
Now we are read to write that one part (the first three terms together) as a square:
I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.
They put it in ( )
I'm going to go ahead and combine those fractions now:
I'm going to factor out a -1 in the second term ( the one in the second ( ) ):
Now I'm going to add (b^2-4ac)/(4a^2) on both sides:
I'm going to square root both sides to rid of the square on the x+b/(2a) part:
Now subtract b/(2a) on both sides:
Combine the fractions (they have the same denominator):