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 Author Topic mattdecamp Posted - 30 March 2006 4:24  Question 48 on the practice test I took has a mistake (simple mistake, but could confuse people not familiar with these kinds of equations)... the question is: "48) If the sum of two numbers is -15 and their product is 56, then these two numbers are roots of which of the following equations?"The correct answer is "x^2 - 15x + 56", but for some reason the answer "x^2 + 15x + 56" is scored as the right answer by the computer... in the explanation it has "(x-7)(x-8)" as the foil, and that turns into "x^2 - 15x + 56", obviously...Anyway, just thought I'd let you guys know. Thisperson Posted - 4 May 2006 13:17  Yes, I was about to post about this; number 48 says:

"If the sum of two numbers is -15 and their product is 56, then these two numbers are roots of which of the following equations?"

It scores the wrong answer as correct, and the last line of the explanation says that:

"Therefore, the roots are x=7 and x=8. "

But clearly, if the roots are 7 and 8, the sum of the roots is 7+8=15, NOT -15 as the question states... (Just to clarify on the post I'm replying to.)

Also, on numbers 30 and 31; the table shows cups sold and hours spent--in the first row, for instance, it says 10 cups were sold and 12 hours were spent. That should mean that the cups of lemonade sold per hour in that trial is 10/12; but if you solve number 30 like that, your answer is wrong. The explanation treats the "Cups Sold" column as cups sold per hour, when that clearly doesn't make sense... In number 31, "Cups Sold" is no longer treated this way and now becomes the number of cups sold per each period of time, to be found in the left column.

If that was too confusing, I'm basically trying to say that 30 and 31 deal with the same data, but 30 interprets it incorrectly... In thirty, it wants the answer of 20 as the median number of cups sold per hour, but 20 is the median number of cups sold per TIME INTERVAL, the lengths of which vary; if you calculate the actual number of cups sold per hour in each trial, not one has a rate of over 3... So the median obviously isn't 20.

Anyway. The wording needs to be changed on this, or the graph, because it doesn't work the way it is. Jeffreyscyler Posted - 15 September 2006 13:29  Omg... i'm a sophmore and i consider my self smart... but wow people freak out big time over a practice test... eluhseia Posted - 29 September 2006 7:45  I also noticed that about th cups per hour I couldn't get an answer from the ones given lalalamia Posted - 27 October 2006 16:44  x^2 + 15x + 56 is the correct answer. The problem says that these 2 numbers are the ROOTS of a certain equation.

From Wikipedia:

"The "root" of a function (f) is the value for x that produces a result of zero ("0")."

So from looking at the problem, we know those two roots are -7 and -8. because they are values of x that make f(x) equal to zero.

The FACTORS of a problem are 7 and 8 as derived from (x-7)(x-8). These are not the roots.

I was confused as to why the answer was that, but then I realized that they were looking for roots, not factors. XD chadhao2005 Posted - 30 October 2006 23:55  we can see
x+y=-15 (a) and x*y=56 (b)
we can change (a) to x=-y-15 (c)
put (c) into (b)
we can get (-y-15)y=56
then -y^2-15y=56
finally we get y^2 + 15y + 56=0

By the way,I'm from China.
Nice to meet u guys! chadhao2005 Posted - 31 October 2006 5:24  i think i'm right........ dizzaa12 Posted - 22 July 2007 9:49  i could be a lil helpful in maths if anybody want me to help,but i also need help in critical readin and essays . thanks Beeblebrox Posted - 31 March 2009 18:4  I had the problems there as well, somehow i managed to get the questions about the cups of lemonade right, but without a multiple choice option, it would have been totally hopeless. Just try something, and hope the result is between the answers...

And besides the problems mattdecamp noted in the first post here (7 + 8 = -15 according to this site), x^2 - 15x + 56 is not even a equation; an equation always contains an equal sign, as the name suggests.

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