Question

If you are solving y 2+2y=48 by completing the square, the next line would be y 2 + 2y - 48 = 0 y 2 + 2y + 4 = 52 y 2 + 2y + 1 = 49

Answer 1

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

`\bf 2(y)[?]=2y\implies [?]=\cfrac{2y}{2y}\implies [?]=1`

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.

`\bf (y^2+2y+1^2-1^2)=48\implies (y^2+2y+1^2)-1=48\implies (y+1)^2=49`

Answer 2

y^2 + 2y = 48.

If we solve this by completing the square, the next line would be:

y^2 + 2y + 1 = 49.

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.