a^3-a^2=48
i can't seem to factor or solve for a. i feel like an idiot! please someone help! i did this like back in high school!
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- rfernandez
- Course Students
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- Joined: Fri Apr 07, 2006 8:25 am
You may very well have learned this in high school, but it's by no means an easy polynomial to factor. It requires the rational roots theorem, which is way out of the scope of the GMAT. I'll spare the details, but here is how the factorization pans out:
a^3 - a^2 = 48
a^3 - a^2 - 48 = 0
*abra cadabra*
(a - 4)(a^2 + 3a +12) = 0
a = 4
(technically, there are two other solutions, but they aren't real numbers... again, out of the scope of the GMAT)
So how would you solve it without algebra? Well, Sudhan just plugged in 4, but he didn't tell us how he knew to plug in 4. Here's one way to get in the right ballpark:
a^3 - a^2 = 48
a^2(a - 1) = 48
Now, a and (a - 1) are very close to each other, so let's just fudge it a little and substitute a for (a - 1).
a^2 * a = 48
a^3 = 48
If a = 3, a^3 = 27. If a = 4, a^3 = 64. So these two numbers (3 and 4) are reasonable guesses to plug in and check because 48 lies between 27 and 64.
Rey
a^3 - a^2 = 48
a^3 - a^2 - 48 = 0
*abra cadabra*
(a - 4)(a^2 + 3a +12) = 0
a = 4
(technically, there are two other solutions, but they aren't real numbers... again, out of the scope of the GMAT)
So how would you solve it without algebra? Well, Sudhan just plugged in 4, but he didn't tell us how he knew to plug in 4. Here's one way to get in the right ballpark:
a^3 - a^2 = 48
a^2(a - 1) = 48
Now, a and (a - 1) are very close to each other, so let's just fudge it a little and substitute a for (a - 1).
a^2 * a = 48
a^3 = 48
If a = 3, a^3 = 27. If a = 4, a^3 = 64. So these two numbers (3 and 4) are reasonable guesses to plug in and check because 48 lies between 27 and 64.
Rey
3 posts Page 1 of 1