math help again
Suppose y is a real number such that 2 < y < 3 and y^3 - 4y - 5 = 0. What is the closest integer to y^2? (no calculator use.... meh)
a. 4 b. 5 c. 6 d. 7 e. 8
please explain why also.... thank you!
November 18, 2012
7 Comments • Newest first
[quote=oAZNvTOFUo]@GreatRomantic WHAAAAAAAAAAAAT[/quote]
Plug and chug. It's you're last resort in every math mc test.
At least you know the right way now.
@GreatRomantic WHAAAAAAAAAAAAT
You can use the Intermediate Value Theorem to find the answer.
It goes like this:
Choose a number between your restriction ---> interval (2,3). Plug it in, see if it is less than or greater than "0".
Now, choose another number between your restriction (2,3), plug it in, and see if it's the OPPOSITE of the sign from the previous.
So if the first time you get GREATER THAN [0], make sure the second time, you get LESS THAN [0].
Thus, if you have both, there exists a number "c", such that f(c) = 0.
You can repeat this process as many times as you wish to "decrease" your desired interval.
You can then just choose "at random" any value within that interval, and that way you can make your interval smaller, to find a possible "c".
And that possible c^2 = one of the multiple choice answers.
And I'm not going to tell you the easier way of finding the answer. MUAHAHAHAHAHAHAHAHA!
can someone just please answer my question?
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i lol'd
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oh, and thank you very much for the answers... however, I needed the explanation also...
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