Hard Quadratic Equations | Algebra

When faced with an unusual or hard quadratic equation problem, 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, difficult quadratic problems have little to do with intuition and a lot to do with understanding how numbers work. If I am looking at a tough 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 equation, 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:

$a + b = -34$

$a\timesb = 5\times24$

Step 1: Prime factorize the product.

$a\timesb = 5\times24 = 2\times2\times2\times3\times5$

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\times). 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.
$2\times2\times2$ and $3\times5$ -> 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 $2\times2\times5\times5$ .

We split it into 2 groups -> $2\times5$ and $2\times5$ (10 and 10). The sum of 10 and 10 is 20.

We split it in another way -> $2\times2$ and $5\times5$ (4 and 25). Their sum is 29. The sum increased.

We split it in yet another way -> 2 and $2\times5\times5$ (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 $2\times2\times2\times3\times5$ and we split it as $2\times2\times2$ and $3\times5$ . 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: $5\times(x - \frac{4}{5})\times(x - \frac{30}{5}) = 0$ (a shortcut to the usual ‘ $5x^2 - 4x - 30x + 24 = 0$ and proceed’ method)

$x = \frac{4}{5}$ 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:

$a + b = -47$

$a\timesb = 8\times(-63) = - 2\times2\times2\times3\times3\times7$

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: $2\times2\times2$ and $3\times3\times7$ 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 $2\times2\times2\times3$ and $3\times7$ . 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 $2\times2\times2\times7$ and $3\times3$ . Now $a$ and $b$ are -56 and 9.
$-56 + 9 = -47$ -> the required sum.

Now the quadratic is $8\times(x + \frac{9}{8})\times(x - \frac{56}{8}) = 0$

$x = \frac{-9}{8}$ or $7$

With a little bit of practice, the hardest quadratic equation 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.

Keep in mind, also, that on the GMAT these difficult quadratic equation problems will always be accompanied by answer choices. Consequently, you generally have the option to test the answer choices to see which would solve the tough quadratic as opposed to slogging through some difficult quadratic formula math. If we revisit Question #2, let’s see it with some answer choices:

Solve for ${x}$ : ${8x^2 - 47x - 63 = 0}$

(A) -12

(B) -7

(D) 12

(E) 21

Here a first step to lighten your load would be to consider negative vs. positive answer choices. Since the first term of the quadratic will be squared, that means that $8x^2$ will be a large positive number regardless of whether $x$ is positive or negative. The second term is subtracted, meaning that if $x$ is positive the $-47x$ term will be negative, but if $x$ is negative the $-47x$ term will be another big positive number. For this reason, you logically don’t have to try choices (A) and (B) – negative answer choices would leave too large a positive value for -63, the third term, to counteract.

From there you can choose which value to try next. Remember: tough quadratic equation problems on the GMAT are often solved via number properties and factors/multiples, so you can think of how the answer choice relate. Two are odd (7 and 21) and one is even (12); two are multiples of 7; and two are multiples of 3. The even/odd relationship can help you eliminate (D) without much extra work: if $x$ were even, then the first two terms would each be even and the last, -63, would be odd. Even – Even – Odd would leave an odd result, but you need the even number 0 to round out this quadratic equation, so your choice is now between (C) and (E).

If you plug in 7, the easier number, note that again you can rely on factoring to solve this difficult quadratic. If $x = 7$ then the quadratic is:

$8(7^2) - 47(7) - 63 = 0$

Every term in this tough quadratic equation is a multiple of 7, so you can factor out your 7s to simplify and check that the quadratic does equal 0:

$7(8 \times 7) - 7(47) - 7(9) = 0$

Divide both sides by 7 and you have:

$56 - 47 - 9 = 0$

This checks out, so (C) is the correct answer to this tough-looking (but not necessarily that tough) quadratic equation problem.

Note how difficult quadratic equation problems on the GMAT work: they’re often testing number properties and number sense much more than they’re testing “tough quadratics,” specifically. So your toolkit to attack hard quadratics is very similar to your math toolkit across the GMAT quant section: be comfortable with factors (including prime factors) and multiples, pay attention to number properties and patterns, and use answer choices to your advantage. And yes you’ll want to practice with quadratics – and in particular tough quadratic equations that don’t include 1 as the coefficient for the a term – but remember that what makes difficult quadratic equation problems solvable on the GMAT is generally a bit more clever – and time-efficient – than simply the quadratic formula.

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