When faced with an unusual quadratic equation, some people waste a lot of time while trying to ‘split the middle term’. The common refrain is ‘I am just not good at it.’ Actually it has little to do with intuition and a lot to do with understanding how numbers work. If I am looking at a quadratic equation and am unable to find the required factors, I will go back to check my quadratic to see if it is correct rather than try to use the esoteric quadratic formula.

To solve a quadratic, you need to find the two factors that add up to give the middle term and multiply to give the product of the constant term and co-efficient of x^2. What you may have problems with is ‘splitting the middle term’. Even though solving a quadratic is a basic skill one must possess to crack the GMAT, we have seen people struggle with it especially if the coefficient of x^2 is something other than 1. Let’s discuss how we can split the middle term quickly in such cases.

Question 1: Solve for x: 5x^2 – 34x + 24 = 0

To factorize, we need to find two numbers a and b such that:

a + b = -34

a*b = 5*24

Step 1: Prime factorize the product.

a*b = 5*24 = 2*2*2*3*5

Step 2: Check the signs and decide what you need. Here the sum of a and b is negative (-34) while product is positive. This means a and b both are negative. Both will add to give a negative number and multiply to give a positive number (more on this at the end*). It also means that both a and b are smaller than 34 (since both are negative, their absolute values will be added to give 34).

Step 3: Try to split the prime numbers into two groups such that their sum is 34. Try the most obvious group first i.e.

2*2*2 and 3*5 -> 8 and 15

If a = 8 and b = 15, we get a + b = 23

But the sum needs to be 34, i.e. a number greater than 23.

Before we discuss the next step, let’s talk about how adding numbers works:

Let’s say the prime factorization we have is 2*2*5*5.

We split it into 2 groups -> 2*5 and 2*5 (10 and 10). The sum of 10 and 10 is 20.

We split it in another way -> 2*2 and 5*5 (4 and 25). Their sum is 29. The sum increased.

We split it in yet another way -> 2 and 2*5*5 (2 and 50). Their sum is 52. The sum increased further.

Notice that further apart the numbers are, the greater is their sum. We get the least sum (20) when the numbers are equal. If we need a higher sum, we increase the distance between the numbers.

Going back to the original question, the prime factorization is 2*2*2*3*5 and we split it as 2*2*2 and 3*5. This gave us a sum of 8 + 15 = 23. We need 34 so we need to get the numbers farther from each other but not too far either. Let’s say, we pick a 2 from 8 and give it to 15. We get two numbers 4 and 30. They are farther apart and their sum is 34. So the numbers we are looking for are -4 and -30 (to get -34 as sum).

Now the quadratic is simply: 5*(x – 4/5)*(x – 30/5) = 0 (a shortcut to the usual ‘5x^2 – 4x – 30x + 24 = 0 and proceed’ method)

x = 4/5 or 6

Question 2: Solve for x: 8x^2 – 47x – 63 = 0

To factorize, we need to find two numbers a and b such that:

a + b = -47

a*b = 8*(-63) = – 2*2*2*3*3*7

The product of a and b is negative which means one of a and b is negative. The sum of a and b is also negative therefore, the number with higher absolute value is negative and the other is positive. When these two numbers will be added, the difference of their absolute values will be the sum and the sign of the sum will be negative. So at least one of a and b is greater than 47. The greater one will be negative and the smaller one will be positive.

Let’s try to split the factors now. To start, we split the primes into two easy groups: 2*2*2 and 3*3*7 to get 8 and 63 but -63 + 8 = -55. We need the sum to have lower absolute value than 55 so we need to get the numbers closer together.

Take off a 3 from 63 and give it to 8 to get 2*2*2*3 and 3*7. Now a and b are -24 and 21; the numbers are too close.

Instead, take off 7 from 63 and give it to 8 to get 2*2*2*7 and 3*3. Now a and b are -56 and 9.

-56 + 9 = -47 -> the required sum.

Now the quadratic is 8*(x + 9/8)*(x – 56/8) = 0

x = -9/8 or 7

With a little bit of practice, the hardest questions can be quickly solved.

*How do you decide the sign of a and b:

Product is positive – This means both a and b have the same sign. If sum is negative, both a and b are negative; if sum is positive, they both are positive.

Product is negative – This means a and b have opposite signs; one is negative, the other is positive. If sum is positive, the number which is positive has a higher absolute value. If sum is negative, the number which is negative has a higher absolute value.

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!*

Hi Karishma,

Obviously another awesome post! The blog post helped me a lot.

Would you say though that this technique would be even faster than simply back-solving when it comes to quadratic equations?

Thanks,

Franco

Hey Franco,

Back solving is very useful if it is possible to use it. Say, the options give you the values of x. Then you can try the options in the quadratic to see what works. But you may not have that luxury. You may need to find the value of x and then do some manipulations on it to get the answer. In that case, it might be too cumbersome to do reverse manipulations on every option and then try them out in the quadratic.

I have got the solution

24*5=2*2*2*3*5

3*5*2+2*2=34

so 30+4=34

therefore 5x^2-34x+24=0

5x^2-30x-4x+24=0

5x(x-6)-4(x-6)=0

(5x-4)(x-6)=0

x=4/5 x=6

yes i did it !

I got the correct answer in question 1 using my teacher’s technique or shortcut (own) about b prime (b’) : ax^2+bx+c=0

-B’ = ± √(B’)^2 – AC all over A

isn’t it (-B±√((B)^2-4AC))/2A