How to solve quadratics when a does NOT equal 1

Question

For example here: running-at-their-respective-constant-rates-machine-x-takes-98599.html
I don't really get how can one quickly factor \(5t^2-34t+24=0\)
If a=1 it is easy a1*a2=c, a1+a2=-b, but what if a=/=1?

Thanks!

jtyee · Answer

Hey dude,

When I do quadratics, I always focus on the "c" portion, in this case the 24. This is helpful because you do not have to even consider the "a", or in this case the 5. Start with the factor pairs for 24 (12,2) (3,8) (4,6). Now think about which, of these three pairs, produces a -34 if you multiply one term by 5 and add it to the other? Also because 24 is (+), both factors need to be either positive or negative. In this case, both will be negative because of -34.

Start with (12,2): hmmm...   + -2? No that doesn't give us -34. What about -12 +  ? No that doesn't give us -34 either
What about (4,6)? Oh look, -4+  gives us -34! Therefore -4 and -6 must be our factor pair

(5t-4)(t-6)

Bunuel · Answer

Check below links. Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm Theory on Algebra: algebra-101576.html Algebra - Tips and hints: algebra-tips-and-hints-175003.html DS Algebra Questions to practice: search.php?search_id=tag&tag_id=29 PS Algebra Questions to practice: search.php?search_id=tag&tag_id=50 Special algebra set: new-algebra-set-149349.html Hope this helps.

sshrivats · Answer

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...

dabral · Answer

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

PareshGmat · Answer

You may use the formula:

ax^2 + bx + c = 0

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

KarishmaB · Answer

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/

How to solve quadratics when a does NOT equal 1

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