Is b less than or equal to 8 in this advanced equation?

[silenzer]

Homework Statement

This isn't going to be this complicated that I have to section my problem down to segments, this is just a part of an advanced problem. There is also something wrong with my browser, I can only type into a textbox where I start typing at the beginning. I can't press enter or anything like that.The problem is this: I cannot understand why this happens: http://j.mp/q2ySH9. I would think that b would be less or equal than BOTH of the numbers... yet it becomes an interval. EDIT: sorry, in the second step it's supposed to be b, not b squared.

Homework Equations

The Attempt at a Solution

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The Attempt at a Solution

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Homework Equations

The Attempt at a Solution

Homework Statement

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The Attempt at a Solution

 

[SammyS]

Homework Statement

This isn't going to be this complicated that I have to section my problem down to segments, this is just a part of an advanced problem. There is also something wrong with my browser, I can only type into a textbox where I start typing at the beginning. I can't press enter or anything like that.The problem is this: I cannot understand why this happens: http://j.mp/q2ySH9. I would think that b would be less or equal than BOTH of the numbers... yet it becomes an interval. EDIT: sorry, in the second step it's supposed to be b, not b squared.
[URL]http://j.mp/q2ySH9[/URL]

(Now your image is visible.)

b^2\le8 ⟶ b^2-(\sqrt{8})^2\le0 ⟶ (b-\sqrt{8})(b+\sqrt{8})\le0

Standard methods give the interval you questioned: -\sqrt{8}\le b\le\sqrt{8}\,.

 

[HallsofIvy]

The "standard methods" are: pq< 0 if and only if p and q are of different sign: either "p< 0 and q> 0" or "p> 0 and q< 0".

Since b^2- 8= (b-\sqrt{8})(b+\sqrt{8}), for that to be less than 0 we must have one of
(a) b- \sqrt{8}&gt; 0 and b+ \sqrt{8}&lt; 0 or
(b) b- \sqrt{8}&lt; 0 and b+ \sqrt{8}&gt; 0

(a) gives b&gt; \sqrt{8}&gt; 0 and b&lt; -\sqrt{8}&lt; 0 but those cannot both be true. Therefore, we must have (b) which gives b&lt; \sqrt{8} and b&gt; -\sqrt{8} or -\sqrt{8}&lt; b&lt; \sqrt{8}.

Since the initial problem had "\le 0" rather than just "<", we must have
-\sqrt{8}\le b\le \sqrt{8}.

By the way, since 8= 4(2) and 4 is a "perfect square", you can write \sqrt{8}= 2\sqrt{2} and that is typically preferred:
-2\sqrt{2}\le b\le 2\sqrt{2}
might be preferred as an answer.