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While solving a quadratic equation in one variable by "splitting the middle term", sometimes I just can't arrive at the two numbers I need and it takes a lot of time.

For eg: $x^2 -124x+1344 = 0$

For solving the above equation by splitting the middle term, I first multiplied the coefficient of $x$ which is $1$ in this case by $1340$ (the constant term). I prime factorised the product, and now I have to pick a group of two numbers such that their sum is $-124$ (coefficient of $x$).

Do we always have to pick random numbers? If only hit and trial works here, what must be our thinking process for picking the other pair of number if the former pair of number doesn't work?

Eg: $x^2 -124x+1344 = 0$

Two numbers from the prime factorization of ($1 \cdot 1344$) is $-64$ and $-21$ but they do not add up to $-124$. What should be my strategy to pick a different pair of numbers?

(I know about the quadratic formula, but it becomes too lengthy sometimes)

leaves
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Harsh Gautam
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  • You seem to have made some mistakes. $-64 \times -21$ is not equal to $1340$, rather $1344$. You need intuition on the sizes of the factors to take a good guess, and also need to recognize when such a factorization is not possible. Indeed, $x^2-124x+1340$ doesn't have an integer root. But for example, $x^2-124x-1340$ does have an integer root, the factorization is $(x-134)(x+10)$. – Sarvesh Ravichandran Iyer Feb 16 '21 at 05:20
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    Yes, this is a hit and trial method only. You better use $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ for $ax^2+bx+c=0.$ – Z Ahmed Feb 16 '21 at 05:21
  • @TeresaLisbon Mam I think I made an error while typing the question it's 1344 instead of 1340 . Can you please elaborate what your last part of comment mean? – Harsh Gautam Feb 16 '21 at 05:24
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    No madam please, only Teresa. See, the point is that not every quadratic can be solved by splitting the middle term, only some can. Do you understand this point? If you must split the middle term, the intuition for what factors can split the middle terms comes from intuition on the "sizes of the factors". For example take $x^2-124x+1344$. You need two numbers which add up to $124$ and whose product is $1344$. The factors are quite far apart from each other, because their sum is quite large but their product is quite small in comparison. You also know that both factors are smaller than $124$. – Sarvesh Ravichandran Iyer Feb 16 '21 at 05:35
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    This intuition can be honed with practice, but the best way to put it is to see how far apart the factors can be , using the sum and product of the factors. It is a little more difficult if the factors have opposite signs, but if they have the same sign it is significantly easier. The answer to the above comes out as $-112$ and $-12$, once you test the smallest factor of $1344$ which is smaller than $124$ i.e. $112$. – Sarvesh Ravichandran Iyer Feb 16 '21 at 05:37
  • @Teresa in the previous comment you have written "The factors are quite far apart from each other, because their sum is quite large but their product is quite small in comparison".But here the sum is-124 and the product is 1344. It can be clearly seen the product is greater than the sum. – Harsh Gautam Feb 16 '21 at 06:20
  • @HarshGautam I was not precise enough, I apologize. For example, take a number, say $70$. If you take two factors of this number which are close to each other, the sum of the two factors is observed to be smaller. For example $14+5$ is bigger than $10+7$ , and $35+2$ is bigger than $14+5$. What I am trying to say, is that the sum and product of the factors tells you intuitively how far apart they are, and using this you can take educated guesses for the middle term split. – Sarvesh Ravichandran Iyer Feb 16 '21 at 06:30
  • @Teresa I think I understood it. Thanks a bunch for this!! – Harsh Gautam Feb 16 '21 at 06:33
  • @HarshGautam Good to know. The best way to learn is examples, you will see plenty of them in MSE or in your book. – Sarvesh Ravichandran Iyer Feb 16 '21 at 06:35
  • @TeresaLisbon What do you mean by "MSE"? – Harsh Gautam Feb 16 '21 at 06:37
  • @HarshGautam Short for "Mathematics Stack Exchange" i.e. this website! – Sarvesh Ravichandran Iyer Feb 16 '21 at 06:46
  • Hint: you can reduce the possibilities by noticing that the sum of two numbers that aggregate to -124 must be both odd or both even [but in this case both odd is impossible. why?]. That leaves us with only five possibilities. – John Joy Feb 16 '21 at 08:18
  • @JohnJoy: There is a way of doing this that doesn't require trial and error (and also doesn't require the quadratic formula). See my answer. – Joe Feb 16 '21 at 09:13
  • @HarshGautam You might want to see the AC method. It is possible to apply this method in reverse to get smaller coefficients, at the cost of the $x^2$ coefficient not being $1$. – Toby Mak Feb 16 '21 at 09:29
  • @Joe: Yes, I'm aware of that .... it's called completing the square, which is what your answer does in a roundabout way. – John Joy Feb 16 '21 at 15:42

5 Answers5

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You need to find two numbers $a$ and $b$ such that \begin{align} a+b &= -124 \\ a \times b &= 1344 \, . \end{align} If you want a way of doing this that doesn't require trial and error, then let $a=-62+k$ and $b=-62-k$ for some constant $k$ (this step involves halving the number that $a$ and $b$ need to add up to, in this case $-124$). Notice that $$ a+b=(-62+k)+(-62-k)=-124 \, . $$ Furthermore, $$ a \times b= (-62+k)(-62-k)=1344 \, , $$ and so \begin{align} 62^2-k^2 &= 1344 \\ 3844-k^2 &= 1344 \\ k^2 &= 2500 \\ k &= 50 \, . \end{align} Hence, $a=-62+50=-12$ and $b=-62-50=-112$. We find that $$ x^2-124x+1344 = (x+a)(x+b)=(x-12)(x-112) $$ and so $x=12$ or $x=112$.

Joe
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  • @TobyMak: The roots need to sum to $124$, but the numbers $a$ and $b$ don't need to, since $(x+a)(x+b)=x^2+(a+b)x+ab=x^2-124x+1344=0$. – Joe Feb 16 '21 at 09:30
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As suggested by Z Ahmed you can use quadratic formula. You will get the two values of x as $$ x = \frac{-(-124)\pm\sqrt{(-124)^2-4(1)(1344)}}{2(1)} $$

After simplification you will get x as $$ x = \frac{124\pm 100}{2} $$

You will get x as 12 and 112

Simplify this and get your answer

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As previously mentioned, you need to find two integers $a,b$ such that $a+b=124, ab = 1344$ due to Vieta's formulas. Now $124$ is divisible by $4$ and $1344$ is divisible by $4^2 = 16$, so we can write $a/4 + b/4 = 124/4, \frac{a}{4} \frac{b}{4} = 1344/4$, and changing variables $m = a/4, n = b/4$ gives $m + n = 31, mn=84$.

If you check the factor pairs of $84$ systematically, for example $(m,n) = (84, 1) = (42, 2)$ and compute their sum, you will find that $m = 28, n = 3$ or vice versa. Thus convert back to the roots by multiplying by $4$ to get $x = 112, 12$.

Toby Mak
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As you know, the factors of $\space 1344\space$ are $\space{1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 24, 28, 32, 42, 48, 56,\\ 64, 84, 96, 112, 168, 192, 224, 336, 448, 672, 1344}$

You should work down the largest factor where $x<124$ because "it" and its cofactor must add up to $124$. It happens on the first try here that $\quad 112+12=124\quad\land\quad 112\times 12=1344\quad$ so there you have it. If you did not find a "pair" by the time the factor became less than the cofactor, then the solution(s) would be rational or irrational, possibly complex, but not integers.

poetasis
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In fact, we can do even better as $1344 = 2^6 \times 3 \times 7$. Dividing $x^2 - 124x + 1344$ by $2^6 = 64$ gives:

$$\frac{x^2}{64} - \frac{124x}{64} + \frac{1344}{64} = 0$$ $$\Rightarrow \left(\frac{x}{8} \right)^2 - \frac{124}{8}\frac{x}{8} + 21 =0 $$ $$\Rightarrow u = x/8: u^2 - \frac{31}{2}u + 21 = 0$$ $$\Rightarrow 2u^2 - 31u + 42 = 0$$

Factor this as $(2u+a)(u+b)$ since $2$ is prime, and the possibilities are limited as $42$ does not have that many divisors. By inspection, this factors as $(2u-3)(u-14) =0 \Rightarrow u = \frac{3}{2}, 14$, and converting back to $x$ with $u = x/8$, $x = 12, 112$.

This method is effectively the AC method in reverse, where we divide the coefficients by a convenient number and use a substitution to obtain a non-monic polynomial.

Toby Mak
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