1. ## Discriminant

Find the smallest value of the interger $\displaystyle b$ for which $\displaystyle 3x^2+bx+2$ is positive for all values of x.

My attempt

since its always postive,

$\displaystyle b^2-4ac<0$
$\displaystyle b^2-4(3)(2)<0$
$\displaystyle b^2-24<0$
$\displaystyle b^2<24$
$\displaystyle b<\sqrt{24}$

however, the answer is $\displaystyle -4$

2. Originally Posted by Punch
Find the smallest value of the interger $\displaystyle b$ for which $\displaystyle 3x^2+bx+2$ is positive for all values of x.

My attempt

since its always postive,

$\displaystyle b^2-4ac<0$
$\displaystyle b^2-4(3)(2)<0$
$\displaystyle b^2-24<0$
$\displaystyle b^2<24$
$\displaystyle b<\sqrt{24}$

however, the answer is $\displaystyle -4$

Well...yes. you forgot that when taking square roots of real numbers we must deal with positive and/or negative values, so

$\displaystyle b^2<24\iff |b|<\sqrt{24}\iff -\sqrt{24}<b<\sqrt{24}\Longrightarrow$ the minimal integer (as required) value b can take is $\displaystyle [-\sqrt{24}]+1=-4$ ...

Tonio

3. Hello, Punch!

And be sure to answer the question.

Find the smallest integer value of $\displaystyle b$ for which $\displaystyle 3x^2+bx+2$ is positive for all values of $\displaystyle x.$

My attempt: .$\displaystyle b^2\:<\:24 \qquad{\color{blue}\Longrightarrow}\qquad {\color{blue}|b| \:<\:\sqrt{24}}$

However, the answer is $\displaystyle -4$

We have: .$\displaystyle -\sqrt{24} \:<\:b\:<\:\sqrt{24}$

. . . . . . . . $\displaystyle -4.9 \:<\:b\:<\:4.9$

So $\displaystyle b$ is between -4.9 and +4.9

Code:
          ↓
- - + o * * * * * * * * * * * * * * * * * o + - -
-5  -4  -3  -2  -1   0   1   2   3   4   5

So the least integer value of $\displaystyle b$ is -4.

Edit: ah, Tonio beat me to it . . . *sigh*
.

4. Thanks! you 2 are a great help heh!