jmuduke08 wrote:
My apologies I meant to type r^2-3r-180=0. Which give the solutions 15 and -12. I just wanted to know the quick way to get these high digit factors. I know how to factor but in terms of quickness, I couldn't think of 15 and -12 without a ton of trial and error which took too long
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bb is right. Use the quadratic formula if it takes a lot of time to proceed by hit and trial.
for hit and trial:
I haven't yet seen any GMAT quadratic question in the form of ax^2+bx+c=0 with fractions/irrational numbers as roots . Almost always the solution will be integers. I'll write my thought process; see if that helps you quicken your hit and trial: (obviously this will look way longer than what happens in my head :) )
r^2-3r-180=0
this is what I make out of 180 and 3 at the first instant:
looking for 2 numbers whose product is 180 and sum is -3.
sum is negative -> implies bigger of the absolute factors is negative and the smaller is positive.
the second step is:
I break 180 into the simplest factors I can see. 18 x 10 in this case.
if it doesn't happen for me (as in this case) i break each into 2 more (simple) factors. (6x3)x(5x2)
this does it for me. I know 12 and 15 are the numbers I'm looking for. The key is not to break 180 into its prime factors but two simpler composite factors.
I don't know whether this helps or not. I just meant to show that you can work on your hit and trial (it should not take a ton of time). Saying that, if you don't have enough time for your GMAT, it would be wise to use the quadratic formula.
Best