Book 1 pg 71 problem 3

[naylor.mike]

Given that 16-y^2=10(4+y), what is y?

16-y^2=10(4+y)
16-y^2=40+10y
-40+10y -40+10y

-y^2-10y-24=0

This is where I got. The answer key shows the next step like this:

16-y^2=40+10y
y^2+10y+24=0

This is where I'm confused--isn't the y^2 negative? Doesn't moving the numbers from the left make them negative as well?

[naylor.mike]

Wait, I think I got it. So instead of subtracting from the right of the =, you subtract from the left. Then you don't have to multiply by -1 . Brilliant!

[tommywallach]

Hey Naylor,

It's not a question of multiplying by negative one. It's that you always solve a quadratic as:

x^2 + bx + c = 0

So if you have -x^2 + bx + c = 0

You need to move everything to the other side:

0 = x^2 - bx - c

Make sense?

-t