# Real roots is defined by...

• Jun 10th 2008, 02:28 PM
theimcq
Real roots is defined by...
Hey everybody,

I'm new here. I'm working on studying and taking practice tests for the College Algebra CLEP test. Basically what CLEP tests do is, if you pass them, you don't end up having to take that class in college (of course, certain colleges accept certain CLEPs as equivalents for their classes, etc, etc).

Anyway, I'm sort of stumped by these real roots of polynomials and crap. Take a look at this question from the practice test:

Quote:

The set of all values of b for which the equation 4x^2 + bx + 1= 0 has either one real root or two real roots is defined by

(A) b > 4
(B) b < 4
(C) b >= 1 or b <= -1
(D) b > 4 or b < -4
(E) b >= 4 or b <= 4 <- CORRECT ANSWER
I'm pretty good with polynomials but I haven't a clue of how they reached this answer. In the shortest of words (if at all possible, lol) how did they reach this answer and conclusion? Could somebody spell it out or direct me to a resource where this concept is demonstrated? Thanks. :)
• Jun 10th 2008, 02:40 PM
o_O
I'm assuming you know the quadratic formula.

The discriminant refers to the expression underneath the square root sign ($\displaystyle b^2 - 4ac$) and is what determines how many roots a quadratic equation (in the form of $\displaystyle ax^2 + bx + c = 0$) has.

1. If $\displaystyle b^2 - 4ac < 0$, then the quadratic has no real roots (can't take the square root of a negative number).

2. If $\displaystyle b^2 - 4ac = 0$, then the quadratic has exactly one real root (only touches the x-axis once).

3. If $\displaystyle b^2 - 4ac > 0$, then the quadratic has 2 real roots.

So, using the above conditions and your quadratic, we have to determine for what values of b gives: $\displaystyle b^{2} - 4(4)(1) \geq 0$

This is a pretty simple inequality so unless you have troubles with solving it, I'll leave the rest to you.
• Jun 10th 2008, 02:44 PM
wingless
$\displaystyle 4x^2 + bx + 1$

Do you know the discriminant?

Discriminant of a polynomial $\displaystyle ax^2 + bx + c$ is, $\displaystyle \Delta = b^2 - 4ac$.

If $\displaystyle \Delta<0$, the polynomial has no real roots.

If $\displaystyle \Delta=0$, the polynomial has one real root. (This root is also referred as a double root)

If $\displaystyle \Delta>0$, the polynomial has two different real roots.

The question asks us to find where the polynomial has one or two real roots. It means $\displaystyle \Delta = 0$ and $\displaystyle \Delta >0$.

So use $\displaystyle b^2 - 4ac \geq 0$

$\displaystyle b^2 - 4\cdot 4 \cdot 1 \geq 0$

$\displaystyle b^2 \geq 16$

It's obvious from here..
• Jun 10th 2008, 02:59 PM
theimcq
Thank you guys so much for both of your very clear explanations! I truly appreciate it. Problem solved! (Clapping)
• Jun 10th 2008, 03:16 PM
masters
Quote:
The set of all values of b for which the equation 4x^2 + bx + 1= 0 has either one real root or two real roots is defined by

(A) b > 4
(B) b < 4
(C) b >= 1 or b <= -1
(D) b > 4 or b < -4
(E) b >= 4 or b <= $\displaystyle \fbox{\color{red} -4}$ <- CORRECT ANSWER Little typo right here. Corrected from original post.
• Jun 10th 2008, 03:17 PM
theimcq
Oops, yup! I copied the problem down incorrectly. Thanks for pointing this out!