Problem Solving

While factorizing a quadratic equation today, i actually took a bit long time to factorize and this amount of time is unacceptable in actual GMAT.

This is the equation - 9x^2 - 155x - 500
It actually took a bit of time to do the next step,
9x^2 - 180x + 25x - 500 = 0
and then proceeded to find the factors.

Is there any quicker way to factorize? As far as i know, i only know this old traditional method to factorize. In GMAT exam, we may face many number of quadratic equations and just cannot afford to spend valuable time in doing this calculation. Please help!

Hey Rapgust ! Could you please post a whole problem.
Well, you could solve this one using quadratic formula. But i don't think answers this big would be present on GMAT.

(-155^2-4*9*500)=D. then x1,2=(-b+-sqrt(D))/2a.
May be question asked to find a sum of 2 roots of the equation ? `

Thanks Nikolayz

That's also another formula i'm aware of. Yes this one can be used but this approach also takes some time to do the calculations stuff.

We could never say anything about what comes in GMAT. I will not be surprised if we get this kind of a big quadratic equation. Better to foresee everything instead of staring at the monitor!

papgust wrote:While factorizing a quadratic equation today, i actually took a bit long time to factorize and this amount of time is unacceptable in actual GMAT.

This is the equation - 9x^2 - 155x - 500
It actually took a bit of time to do the next step,
9x^2 - 180x + 25x - 500 = 0
and then proceeded to find the factors.

Is there any quicker way to factorize? As far as i know, i only know this old traditional method to factorize. In GMAT exam, we may face many number of quadratic equations and just cannot afford to spend valuable time in doing this calculation. Please help!

Were you just factoring the quadratic for fun? I thought Ian was the only person who did that!

It's extremely rare to see a quadratic on the GMAT with a coefficient other than 1 for the squared term.

When you do see such a problem, there's always a "trick" to solving without using the quadratic equation - I explicitly tell all my students NOT to bother memorizing the formula, since you'll never have to use it (although there may be questions on which you can use it).

For example, if this equation showed up in problem solving and we were asked to find a solution, we could simply backsolve (plug in the choices).

In other cases, the coefficient can be factored out.

e.g. "If 3x^2 + 27x - 30 = 0, what's one possible value for x?"

We could just rewrite the quadratic as:

3(x^2 + 9x - 10) = 0

and solve as usual.

In another question I saw, you could use part of the question to help you quickly solve. It was something like:

Which of the following is equivalent to (6x^2 + 19x + 15)/(3x + 5)?

and all of the choices were simple binomials, so you knew that (3x+5) had to be one of the solutions, making it trivial to find the other solution (6x^2/3x = 2x and 15/5 = 3, so you knew the other binomial had to be (2x+3)).

Even if you didn't see that trick, you could have picked numbers to solve instead of doing algebra.

If the quadratic you posted is from an actual question (and not from the 4th edition of "Fun Quadratics for long Car Trips"), please post the whole thing!

hehe Nice post Stuart =)
+1 for the whole problem

Hi Stuart,

Thanks for your tricks to solve quadratic equations. I never tried to solve this equation for fun like Ian used to I'm currently going through my basics in high school math book and while trying to solve a problem, i faced this equation. Glad to hear that this kind of a quadratic equation is a rare case.

Stuart Kovinsky wrote: Which of the following is equivalent to (6x^2 + 19x + 15)/(3x + 5)?

and all of the choices were simple binomials, so you knew that (3x+5) had to be one of the solutions, making it trivial to find the other solution (6x^2/3x = 2x and 15/5 = 3, so you knew the other binomial had to be (2x+3)).

The example you pointed out already have a factor of (3x+5). So if one factor is already given, the other factor can be found using this method.

But any tricks on how to find all the factors for a quadratic equation i posted where a common number cannot be taken out like your first example?

papgust wrote:Hi Stuart,

The example you pointed out already have a factor of (3x+5). So if one factor is already given, the other factor can be found using this method.

But any tricks on how to find all the factors for a quadratic equation i posted where a common number cannot be taken out like your first example?

As I said, it's extremely unlikely you'd have to do so on the GMAT.

However, you can start by looking at the first and last terms. In the example I gave, we have 6x^2 and 15. 15 has a limited number of possible factor pairs (1*15 or 3*5), as does 6 (1*6 or 2*3), so we can use trial and error effectively to find the pairs that also generate the right number of "x"s in the middle term.

If you actually needed to solve this for a math test you're probably also getting marks for your solution, so the quadratic formula would be appropriate; fortunately we don't care about getting 10/10 on the GMAT, just 1/1.