How to solve quadratics when a does NOT equal 1

I think the speed with factorization comes with practice.. you have to be using numbers, especially multiplication tables so often that they pop into your mind. I don't think there can be a 'method' to do it 'quickly'

for 5(t^2)-34t+24 .. I just multiply a*c --> 5*24 = 120.

now of course you could try to factorize 120 and list all its factors.. thats the 'method'. But the moment I see 120 and -34, I know the second term should be split into -30 and -4 ( 30 * 4 =120 pops into my mind immediately)

5(t^2) -30t -4t +24 = 5t( t-6) -4(t-6) = (t-6) (5t-4)..

Of course this gets more difficult when dealing with bigger numbers or multiplications we use less often...

I would recommend using the method that @sshrivats described above. Generally, on the GMAT the quadratics that have a coefficient other than 1 can be solved by reasonably guessing the factors. They will not give you a quadratic that is a pain to factor.

Another example of such a quadratic from an official GMAT equation is: 2x^2 - 5x - 33 = 0, here the product of 2 and (-33) is -66, the two numbers that give the product of 66 are 11 and 6(they also give a add up to 5 if we choose the right signs), we have to choose the signs based on the two numbers adding up to -5(the coefficient of x). In this case the two numbers are -11 and 6. The quadratic equation then becomes:

2x^2 - 11x + 6x - 33 = 0

x(2x-11) + 3(2x-11)=0

(x+3)(2x-11)=0

Cheers,
Dabral

You may use the formula:

\(ax^2 + bx + c = 0\)

\(Roots = \frac{-b + - \sqrt{b^2 - 4ac}}{2a}\)

In one of my weekly posts, I have taken exactly this question: "How to factorize unusual quadratic equations?" The first example chosen is this one.
So you should check it out: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/1 ... equations/