You could simplify this work by factoring "3" out of all four terms, as follows:
3(x^2 + 2x - 3) =3(0) = 0
Hold the 3 for later re-insertion. Focus on "completing the square" of x^2 + 2x - 3.
1. Take the coefficient (2) of x and halve it: 2 divided by 2 is 1
2. Square this result: 1^2 = 1
3. Add this result (1) to x^2 + 2x, holding the "-3" for later:
x^2 +2x
4 Subtract (1) from x^2 + 2x + 1: x^2 + 2x + 1 -3 -1 = 0,
or x^2 + 2x + 1 - 4 = 0
5. Simplify, remembering that x^2 + 2x + 1 is a perfect square:
(x+1)^2 - 4 = 0
We have "completed the square." We can stop here. or, we could solve for x: one way would be to factor the left side:
[(x+1)-2][(x+1)+2]=0 The solutions would then be:
x+1-2=0=> x-1=0, or x=1, and
x+1 +2 = 0 => x+3=0, or x=-3. (you were not asked to do this).
Answer:
answer is a
Step-by-step explanation:
Answer:
10-11 is -1
Step-by-step explanation:
i don'T care much for attachments ;)
-3x^2-2x (+-)6=0
ax^2+bx+c
a=-3
b=-2
c= either positive or negative 6 (there is no plus or minus in your question.)
What are the answer choices? It might help.
6 and 8 I believe is the answer
