Factoring Quadratic Equations efficiently.

Question

Hi Guys,

With the equation like this, if we are looking to find the value of N, we can always prime factorize and test different combinations to see what would give us 1890 when multiplied and 3 when subtracted, but I find with bigger numbers this process can take a bit of time as it requires some trial and error. I wondered if there is a more "efficient" approach when dealing with bigger numbers like this, that would quickly tell us which factors are to be used?

N^2+3n-1890

Thanks

GMATinsight · Answer

I deal with them as follows N^2+3n-1890 I.e. N^2+3n = 1890 I.e. N(N+3) = 1890 I.e. Product of two numbers separated by 3 resulting in 1890 Also, One of them must be a multiple of 5 because result is divisible by 5 Let's take approximate square root of 1890 which is approx 42 so the possibility may be 42*45 which fits well So N may be 42 or -45 I hope this helps!!!

rheam25 · Answer

I found this to be useful for solving equations with coefficients, just wanted to share it incase anyone is struggling with this like me

https://www.youtube.com/watch?v=kqW8RjcaXEc

TargetMBA007 · Answer

That's a cool strategy. Did you find a lot of questions on the GMAT with a co-efficient on  x^2 ?

rheam25 · Answer

I haven’t seen too many questions directly asking you to factor it with a coefficient but yes it does come in handy with inequalities

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Factoring Quadratic Equations efficiently.

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