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I have a problem that says:
Suppose $3x^2+bx+7 > 0$ for every number $x$, Show that $|b|<2\sqrt21$.

Since the quadratic is greater than 0, I assume that there are no real solutions since
$y = 3x^2+bx+7$, and $3x^2+bx+7 > 0$, $y > 0$
since $y>0$ there are no x-intercepts. I would use the discriminant $b^2-4ac<0$.

I now have $b^2-4(3)(7)<0$
$b^2-84<0$
$b^2<84$
$b<\pm\sqrt{84}$

Now how do I change $b$ to $|b|$? Can I take the absolute value of both sides of the equation or is there a proper way to do this?

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If you have an equation, you may take the absolute value of both sides. But what you have here is not an equation. –  wj32 Sep 23 '12 at 1:15
    
If you want to type $\sqrt{21}$ instead of $\sqrt21$, code it like this: \sqrt{21}. –  Michael Hardy Sep 23 '12 at 1:23
    
@MichaelHardy Thank's for the tip! I was trying to figure that out. –  Kot Sep 23 '12 at 1:28

4 Answers 4

up vote 3 down vote accepted

From $b^2\lt 84$, you cannot conclude that $b\lt \pm\sqrt{84}$, whatever that may mean. It cannot mean that $b\lt \sqrt{84}$ or $b\lt -\sqrt{84}$, since $-100$ is less than each of $\sqrt{84}$ and $-\sqrt{84}$.

What you probably intend to say is that $b^2\lt 84$ iff $-\sqrt{84}\lt b\lt \sqrt{84}$. And we can rewrite this double inequality as $|b|\lt\sqrt{84}$.

More simply, note that for any $b$, we have $\sqrt{b^2}=|b|$. To check this is true, verify it holds when $b\ge 0$ and when $b\lt 0$.

So we can conclude directly from $b^2\lt 84$ that $|b|\lt \sqrt{84}$. And $\sqrt{84}=2\sqrt{21}$.

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Is $\sqrt{b^2} = |b|$ an identity or property? How come I have never learned that before? –  Kot Sep 23 '12 at 1:38
1  
Well, it is true for all reals $b$, so I guess it is an identity. Easy to establish, occasionally useful. Be careful about algebraic manipulation of inequalities, mistakes are too easy to make. I prefer in general to see what's going on, by visualizing where the relevant things must be on a "number line." –  André Nicolas Sep 23 '12 at 1:43

Use the result that $a^2>b^2$ if and only if $|a|>|b|$.

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We have $-3<2$ yet $|-3|<|2|$ is false. If $b^2<84$ then $b^2-84<0$ from where $$(b-\sqrt{84})(b+\sqrt{84})<0$$

Now, $a\cdot b<0$ only if $a,b<0$ or $a,b>0$.

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What you've written is an inequality, not an equation. If you have an equation, say $a=b$, you can conclude that $|a|=|b|$.

But notice that $3>-5$, although $|3|\not>|-5|$.

If $3x^2+bx+7>0$ for every value of $x$, then the quadratic equation $3x^2+bx+7=0$ has no solutions that are real numbers. THat implies that the discriminant $b^2-4ac=b^2-4\cdot3\cdot7$ is negative. If $b^2-84<0$ then $b^2<84$, so $|b|<\sqrt{84}$.

Now observe that $\sqrt{84}=\sqrt{4}\sqrt{21}=2\sqrt{21}$.

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