# How do you misinterpret homework/quiz/test questions?

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Comp_Geek_573
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02 Nov 2012, 12:39 pm

I usually do well on tests, but one question springs to mind for me:

(True/False)
In the following theorem, x is a free variable:
Suppose a, b and c are positive real numbers and b^2 > 4ac. Then the equation ax^2 + bx + c = 0 always has two solutions.

Under the time pressure of the pop quiz this was asked in, I thought it was asking whether the THEOREM was true or false, not whether x was a free variable.

The class is called "Foundations of Digital Computing" and is about logical proofs - it is NOT an algebra class. You, in fact, need algebra to take this course.

How do you tend to misinterpret quiz/test questions?

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ianorlin
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02 Nov 2012, 1:24 pm

I actually think your interpretation is right.

The way to say what the professor meant is In the following theorem is x a free varible?

antifeministfrills
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02 Nov 2012, 3:51 pm

You got the right interpretation, and the answer is true, I think.

profofhumanities
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02 Nov 2012, 3:54 pm

Oh, mercy. It's high end math. I would have been at a complete loss. A free variable isn't incarcerated?

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Comp_Geek_573
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02 Nov 2012, 6:31 pm

I answered true. I got it wrong because the class was discussing "free" versus "dummy" variables and it was, in fact, asking whether x was a free variable. It wasn't.

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littlelily613
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03 Nov 2012, 12:47 pm

This is why I am not a math major.....

I would employ the use of "Eeny Meeny Miny Moe" here....

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profofhumanities
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03 Nov 2012, 1:01 pm

littlelily613 wrote:
This is why I am not a math major.....

I would employ the use of "Eeny Meeny Miny Moe" here....

That was funny, and I agree. My BSE and MA are in English for one reason: no math.

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Ancalagon
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04 Nov 2012, 12:23 am

Comp_Geek_573 wrote:
(True/False)
In the following theorem, x is a free variable:
Suppose a, b and c are positive real numbers and b^2 > 4ac. Then the equation ax^2 + bx + c = 0 always has two solutions.

Let a=c=1, b=2
ax^2 + bx +c = (x+1)^2, which has only one solution, x=-1

So it would have been false anyway. Personally I think the question was worded poorly.

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ianorlin
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04 Nov 2012, 11:18 am

Ancalagon wrote:
Comp_Geek_573 wrote:
(True/False)
In the following theorem, x is a free variable:
Suppose a, b and c are positive real numbers and b^2 > 4ac. Then the equation ax^2 + bx + c = 0 always has two solutions.

Let a=c=1, b=2
ax^2 + bx +c = (x+1)^2, which has only one solution, x=-1

So it would have been false anyway. Personally I think the question was worded poorly.

It says b^2>4aC but 2^2=41(1) =4 If it was > it would be false