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Intern
Joined: 22 Apr 2016
Posts: 16

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11 Jun 2016, 04:40
Hi,

If you have "ax²+bx+c=0" and if you can't find two numbers (c and d) whose sum is equal to "b" and whose product is equal to "c", does it mean that you can't factorize "ax²+bx+c" ?

And if you can't factorize "ax²+bx+c", does it mean that "ax²+bc+c=0" has no solution ?

Thanks a lot !
Math Expert
Joined: 02 Aug 2009
Posts: 8601

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11 Jun 2016, 04:59
jbyx78 wrote:
Hi,

If you have "ax²+bx+c=0" and if you can't find two numbers (c and d) whose sum is equal to "b" and whose product is equal to "c", does it mean that you can't factorize "ax²+bx+c" ?

And if you can't factorize "ax²+bx+c", does it mean that "ax²+bc+c=0" has no solution ?

Thanks a lot !

Hi,

In normal scenario, a QUADRATIC equation will have two roots..
But it is possible these roots are imaginary number in some case...

Since GMAT does not deal with Imaginary number, there will be no solution for such equation as per GMAT..
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Joined: 04 Dec 2015
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GMAT 1: 790 Q51 V49
GRE 1: Q170 V170

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14 Jun 2016, 14:06
1
jbyx78 wrote:
Hi,

If you have "ax²+bx+c=0" and if you can't find two numbers (c and d) whose sum is equal to "b" and whose product is equal to "c", does it mean that you can't factorize "ax²+bx+c" ?

And if you can't factorize "ax²+bx+c", does it mean that "ax²+bc+c=0" has no solution ?

Thanks a lot !

It depends on what you mean by 'not being able to find' those two numbers! Some quadratics include numbers that are incredibly tough to factor - for instance, x² + 2x - 399 = 0. I wouldn't want to be in the position of finding two numbers that multiply to -399, and add to 2. Nonetheless, those numbers exist - that quadratic factors to (x + 21)(x - 19) = 0. It would be easy to see a problem like that, and assume, after trying for a while, that those numbers didn't exist. Be careful about making that assumption.

If you see a quadratic that you're struggling to use this technique on, there's probably something else you could do. Try the quadratic formula. Or, try going back in the problem and seeing if you could simplify it differently. In the example I just gave, this is a much easier solution (the example, by the way, is actually adapted slightly from an Official Guide problem):

x² + 2x - 399 = 0
x² + 2x + 1 = 400
(x + 1)(x + 1) = 400
x + 1 = +/- 20
x = -21 or +19

chetan2u is right, too. There are some quadratics for which a solution just doesn't exist. And some quadratics only have one solution. You'll see the latter on the GMAT sometimes, but I'm not sure that I've ever seen the former.

Here's an example of a quadratic with no real solutions:

x² + x + 4 = 0

And here's an example of a quadratic with one solution:

x² + 4x + 4 = 0
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Joined: 29 Sep 2014
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14 Jun 2016, 15:26
Good comments above, but also there is an easy way to check whether a quadratic equation has real roots using the quadratic formula.

$$x = (-b +/- \sqrt{b^2-4ac})/2a$$

If you take just the bit of the quadratic formula underneath the square root sign , you can use that to see if the equation has zero, one or two real roots.

If $$b^2-4ac$$ is negative, then the equation has no real roots, if it is 0 then the equation has one root, and if it is positive then the equation has two real roots!