so let's start off by grouping y²+2y=48
(y² + 2y +[?]²) = 48
now, we have a missing value to get a perfect square trinomial.
recall that the middle term is the product of 2 and the other two without the ², so one can say that
[tex]\bf 2(y)[?]=2y\implies [?]=\cfrac{2y}{2y}\implies [?]=1[/tex]
low and behold, is 1, now, since all we're doing is borrowing from our very good friend Mr Zero, 0, if we add 1², we also have to subtract 1².
[tex]\bf (y^2+2y+1^2-1^2)=48\implies (y^2+2y+1^2)-1=48\implies (y+1)^2=49[/tex]
Answer:
y^2 + 2y = 48.
If we solve this by completing the square, the next line would be:
y^2 + 2y + 1 = 49.
Step-by-step explanation:
To create a trinomial square on the equation's left side, we must first find a value that is equal to the square of half of b.
(b/2)^2 = (1)^2
Add the term to each side.
y^2 + 2y + (1)^2 = 48 + (1)^2
Simplify it.
y^2 + 2y + 1 = 49.
