College Algebra CLEP Exam Sample Question #2

F

FamilyGuy0395

New member
Joined
Sep 29, 2019
Messages
16
Question: Let a, b, and c be real numbers, where a ≠ 0. If the equation ax^2 + bx + c = 0 has two real solutions, which of the following statements could be true? Indicate all such statements.

Choices:
A. a > 0, b > 0, and c < 0
B. b = ±√ac
C. ac < 0

Correct Answers: A & C

Can anyone help explain to me how to even begin this problem? I know what the correct answers are, but even that still doesn't help me begin to understand how to begin to solve a problem like this.
 
If a quadratic has two real roots, then its discriminant must be positive:

[MATH]b^2-4ac>0[/MATH]
Can you see that choices A and C make the discriminant positive? Can you see that choice B can make it negative (if \(ac>0\))?
 
@MarkFL

For choice A, I set a = 1, b = 1, and c = -1, and found the discriminant to be 5 which is > 0, so I know that A must be true.

However, I'm having a hard time doing this same thing with choices B and C to determine if they are true or not.

Is there some easier way of testing each of these answers without having to come up with theoretical sets of numbers? I feel like this question would take too long if done my way.
 
FamilyGuy0395 said:
@MarkFL

For choice A, I set a = 1, b = 1, and c = -1, and found the discriminant to be 5 which is > 0, so I know that A must be true.

However, I'm having a hard time doing this same thing with choices B and C to determine if they are true or not.

Is there some easier way of testing each of these answers without having to come up with theoretical sets of numbers? I feel like this question would take too long if done my way.
Click to expand...
Your method is pretty dangerous. It worked for 1, 1, -1. But what about 1, 1, -2? Or 2, 1, -35? Etc. You need to formally prove that if a > 0, b > 0, and c < 0 then b2−4ac>0.
 
@lev888 If I'm thinking about this correctly, b^2 is always positive, and -4ac, when one variable > 0 and one is < 0, will always be positive.
 
FamilyGuy0395 said:
@MarkFL

For choice A, I set a = 1, b = 1, and c = -1, and found the discriminant to be 5 which is > 0, so I know that A must be true.

However, I'm having a hard time doing this same thing with choices B and C to determine if they are true or not.

Is there some easier way of testing each of these answers without having to come up with theoretical sets of numbers? I feel like this question would take too long if done my way.
Click to expand...
You cannot choose numbers to prove something true. You can choose numbers to prove something false.

[MATH]\text {two real roots} \iff b^2 - 4ac > 0 \implies 2 \text { real roots} \iff b^2 > 4ac.[/MATH]
[MATH]b > 0 \implies b^2 > 0.[/MATH]
[MATH]a > 0 \text { and } c < 0 \implies ac < 0 \implies 4ac < 0 \implies b^2 > 4ac \implies 2 \text { real roots.}[/MATH]
So A is true. Now see what that tells you about C.

[MATH]b = \pm \sqrt {ac} \implies ac \ge 0 \text { and } b^2 = ac \le 4ac \implies[/MATH]
[MATH]b^2 \not > 4ac \implies \text {number of real roots} \ne 2.[/MATH]
So B is false.

This is primarily about reasoning with "if and only if" contingencies.
 
@lev888 I think I can make out C. Like we said previously, b^2 will always be positive, and if ac < 0, the term 4ac will always be 0, which will end up being a positive number minus 0, always leading to a positive number.
 
FamilyGuy0395 said:
@lev888 I think I can make out C. Like we said previously, b^2 will always be positive, and if ac < 0, the term 4ac will always be 0, which will end up being a positive number minus 0, always leading to a positive number.
Click to expand...
No, b^2 is not always positive, it can be 0 - see my post #8. Now we don't have the b>0 condition, so b^2 is greater than, or equal to 0.
Can you explain why 4ac is 0?
 
FamilyGuy0395 said:
Question: Let a, b, and c be real numbers, where a ≠ 0. If the equation ax^2 + bx + c = 0 has two real solutions, which of the following statements could be true? Indicate all such statements.

Choices:
A. a > 0, b > 0, and c < 0
B. b = ±√ac
C. ac < 0

Correct Answers: A & C

Can anyone help explain to me how to even begin this problem? I know what the correct answers are, but even that still doesn't help me begin to understand how to begin to solve a problem like this.
Click to expand...
"Let a, b, and c be real numbers, where a ≠ 0. If the equation ax^2 + bx + c = 0 has two real solutions, which..."

If the case of two "equal" solutions is excluded, it should have been explicitly stated (in my opinion). This is one of the reasons, I cannot "make-up" multiple-choice-questions for a test.
 
@lev888 You're right, I forgot that we don't have that condition for b anymore. Thus, b can be > or = 0. If ac < 0, that means 4ac will always be negative, and b^2 (which is either 0 or > 0) will subtract a negative number, always leading to a positive number, which is why option C is true. Is that correct? If so, can you help explain why B is not true?
 
FamilyGuy0395 said:
Question: Let a, b, and c be real numbers, where a ≠ 0. If the equation ax^2 + bx + c = 0 has two real solutions, which of the following statements could be true? Indicate all such statements.

Choices:
A. a > 0, b > 0, and c < 0
B. b = ±√ac
C. ac < 0

Correct Answers: A & C

Can anyone help explain to me how to even begin this problem? I know what the correct answers are, but even that still doesn't help me begin to understand how to begin to solve a problem like this.
Click to expand...
I think everyone is misreading the problem (except for the OP, initially).

It doesn't ask which choices imply that there are two [distinct] real solutions, but which are compatible with two real solutions. That is, it is looking neither for necessary or sufficient conditions, but for conditions that permit two real solutions in some cases.

A could be true, because an example has been found in which there are two real solutions, namely x^2 + x - 1 = 0, where the discriminant is 5.

C could be true, using the very same example.

How about B? Well, if b = ±√(ac), then b^2 = ac > 0, and b^2 - 4ac = b^2 - 4b^2 = -3b^2 < 0, so there can't be any real solutions.

So if there are two real solutions, then it may be true that a > 0, b > 0, and c < 0, and it may be true that ac < 0, but it can't be true that b = ±√(ac).

Am I misreading the problem?
 
FamilyGuy0395 said:
@lev888 You're right, I forgot that we don't have that condition for b anymore. Thus, b can be > or = 0. If ac < 0, that means 4ac will always be negative, and b^2 (which is either 0 or > 0) will subtract a negative number, always leading to a positive number, which is why option C is true. Is that correct?
Click to expand...
Yes

FamilyGuy0395 said:
If so, can you help explain why B is not true?
Click to expand...
See posts 11 and 16.
 
I think through all of your responses I can put it together why A & C are true and B isn't:

A. a > 0, b > 0, & c < 0

b^2 - 4ac > 0
(+) - 4(+)(-) > 0
(+) + (+) > 0

B. b = ±√ac

b^2 - 4ac > 0
(√ac)^2 - 4ac > 0
ac - 4ac > 0
ac(1 - 4) > 0
ac(-3) > 0
-3ac NOT > 0

C. ac < 0

b^2 - 4ac > 0
0/(+) - 4(-) > 0
0/(+) + (+) > 0

Can you understand my logic, @lev888? If so, does this look right? By the way, what I mean by (+) and (-) are any positive or any negative number.
 
You must log in or register to reply here.
Top