ok...im confused with the expression: Y^2 < 64.

Wouldn't this be square rooted to Y < +or - 8?

Why does the book say - 8 < y < 8?

Is it because you have to consider that Y might be negative and the sign would then switch when you divide by the negative?

thank you

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I think it should be

+- Y < 8, so then you have

+Y < 8 which is fine,

and

-Y < 8.
Divide by -1 to get Y > -8

So the answer is -8 < Y < 8

If y^2 < 64

Remember that a variable squared is tricky because it hides the true sign of the number. If the problem had an = sign you would say Y = 8 and Y = -8. The same goes for the inequality. We're going to have two answers for this problem and will express our "AND" differently.

So here are our two functions:

1) Y^2 < 64

AND

2) -Y^2 < 64

Solve:

1) Y^2 < 64
Square root of both sides gives...

Y < 8

2) -Y^2 < 64
Square root

-Y < 8

Divide by -1

Y > -8


So we have our two equations and combine them. Remember the "and"? Just rewrite them and combine.

Y<8
and
Y>-8 --> -8<Y

-8<Y<8

Here's a general method you can use to solve many polynomial inequalities:

y^2 < 64
y^2 - 64 < 0
(y + 8) (y - 8) < 0

So you're interested in values of y that make the left side of the inequality take on a negative value. We'll start with the numbers y=-8 and y=8. We know that the left side of the inequality equals zero for those two values of y and for no others. So we can test the values of y outside of and between those special numbers -8 and 8 to see if they will yield a positive or a negative value:

1) y<-8

Plug in a number smaller than -8 into (y + 8) (y - 8) and what happens? You get a negative times a negative --> positive

2) -8<y<8

Plug in a number between -8 and 8 into (y + 8) (y - 8) and what happens? You get a positive times a negative --> negative

3) y>8

Plug in a number greater than 8 into (y + 8) (y - 8) and what happens? You get a positive times a positive --> positive

So the only range that solves the inequality is -8<y<8.

Rey