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15 May 2020, 06:00
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Hi,
I know that to factorize a quadratic equation, we need to find a pair of numbers which should be the product of the last term and also add up to the number in the middle term. However for large numbers how can we systematically find such a pair which will satisfy both requirements?
I know that to factorize a quadratic equation, we need to find a pair of numbers which should be the product of the last term and also add up to the number in the middle term. However for large numbers how can we systematically find such a pair which will satisfy both requirements?
27 May 2020, 07:17
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rajak01
It's important, when learning algebra, to distinguish carefully between an "expression" and an "equation". An expression is just an arithmetic combination of letters and numbers. So 29, 7x, 3x + 11, x^5 - 0.5 and x/9 are five different simple examples of expressions. If you see an expression alone, you have no idea what it is equal to. The only thing you're legally allowed to do with an expression is to rewrite it in ways that do not change its value. So if you saw the expression 3x/6, it would be fine to rewrite that as x/2, since by cancelling the '3' we are not changing the value of the expression. But that's the only useful thing we can do with "3x/6".
An equation tells you that one expression has the same value as another expression. When you have an equation, you can do a ton of things - you can rewrite equations in all kinds of ways, provided you're doing the same thing to both sides (with some minor caveats).
A "quadratic" is a type of expression, one that contains a letter that is squared, often added to a letter to the first power and to a number. The letter terms might be multiplied by a number too. So x^2, or x^2 + 5x + 6 or x^2 - 16 are simple examples of quadratics. If you only see a quadratic on its own, then you're only looking at an expression. You have no idea what it's equal to, so you can't write down any kind of equation - that would always be a serious mathematical mistake. Do not assume that a quadratic expression is equal to zero. You can, however, rewrite a quadratic in equivalent ways, especially by factoring (that's almost always going to be useful). So if you saw the quadratic x^2 - 16, a difference of squares, you could rewrite that as (x + 4)(x - 4).
A "quadratic equation" is an equation that is written, or that can be rewritten, so that it has a quadratic expression on one side and zero on the other. So x^2 = 16, or x^2 - 16 = 0, or x^2 + 5x + 6 = 0 are examples of quadratic equations. When we know that our quadratic is equal to zero, then we can (at least for examples as simple as those on the GMAT) use the familiar quadratic factoring technique to find the solution or solutions for x.
In the question you link to above, the quadratic is not equal to zero. We know that t^2 - kt - 48 has (t - 8) as a factor. The quadratic must have a second factor, which will look like (t + a), where a is some number. So
t^2 - kt - 48 = (t - 8)(t + a)
If you multiply out the right side, you get t^2 + (a - 8)t - 8a. Here (a - 8) and 8a are just numbers, and number at the end on the left side, "-48", must equal the number on the end on the right side, "-8a". So a = 6. We want to find k, and since -k is in front of the "t" on the left side, that must equal (a-8), the number in front of the "t" on the right side. Since a = 6, we know -k = a - 8 = 6 - 8 = -2, and if -k = -2, k = 2.
That's a bit complicated, and some solutions in the thread you link to make use of an important fact. If (t - 8) is a factor of a certain quadratic *expression*, that always means that *if* you set that expression equal to zero, then t = 8 will be a solution of the equation you get. So the quadratic is not inherently equal to zero; instead we're saying "if the quadratic were equal to zero, then 8 would be a solution for t" and drawing conclusions about k from that.
General Discussion
20 May 2020, 04:32
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In a quadratic, the more factors the number at the end has, the longer it typically takes to find a factorization. Since you only have two minutes per question on the GMAT, there's a limit to how complicated quadratics will get, so you don't need to be concerned about really crazy examples. But if you had something like this, which I think is still more complicated than anything I've seen on the real test:
x^2 + 74x + 1200 = 0
If you guess a pair of numbers that multiply to 1200 -- say you guess 20 and 60 at first -- and their sum is too large, then you can know that you want two numbers that are closer together; the closer the two numbers are in value, the smaller their sum will be. So you then might guess 30 and 40. These now have a sum that is too small, so we want two numbers further apart than that. So now we know we want one number between 20 and 30, and another between 40 and 60. Since we need a number between 20 and 30 that is also a factor of 1200, then, using divisibility principles, the only candidates are 24 and 25 (we can rule out numbers like 21 and 28 because 1200 is not divisible by 7, for example, and similarly can rule out every other number between 20 and 30). If the sum of the two numbers is 74, one number can't be 25, because then both numbers would be odd and would not multiply to 1200, so the two numbers must be 24 and 50 (note that you can find the 50 using the sum of 74 rather than by using any multiplication or division).
The more you know about number theory -- divisibility especially -- the faster this becomes. There are other systematic ways to do this kind of thing, for example by finding the prime factorization of the number at the end to more easily identify its divisors, but those are needlessly time-consuming in most cases.
x^2 + 74x + 1200 = 0
If you guess a pair of numbers that multiply to 1200 -- say you guess 20 and 60 at first -- and their sum is too large, then you can know that you want two numbers that are closer together; the closer the two numbers are in value, the smaller their sum will be. So you then might guess 30 and 40. These now have a sum that is too small, so we want two numbers further apart than that. So now we know we want one number between 20 and 30, and another between 40 and 60. Since we need a number between 20 and 30 that is also a factor of 1200, then, using divisibility principles, the only candidates are 24 and 25 (we can rule out numbers like 21 and 28 because 1200 is not divisible by 7, for example, and similarly can rule out every other number between 20 and 30). If the sum of the two numbers is 74, one number can't be 25, because then both numbers would be odd and would not multiply to 1200, so the two numbers must be 24 and 50 (note that you can find the 50 using the sum of 74 rather than by using any multiplication or division).
The more you know about number theory -- divisibility especially -- the faster this becomes. There are other systematic ways to do this kind of thing, for example by finding the prime factorization of the number at the end to more easily identify its divisors, but those are needlessly time-consuming in most cases.
