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?
$\endgroup$ 34 Answers
$\begingroup$Use the result that $a^2>b^2$ if and only if $|a|>|b|$.
$\endgroup$ $\begingroup$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 $\sqrt{84}=2\sqrt{21}$.
$\endgroup$ 2 $\begingroup$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 $a,b>0$.
$\endgroup$ $\begingroup$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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