How to solve Quadratic Equation in less time

Quadratic equations are considered to be one of the most complex and tricky parts of any exam having a section Quantitative Aptitude. It is seen that students, generally, tend to make mistakes and at times take more time in solving questions. Considering the time available in any competitive exam, it is important that the question on Quadratic Equations are solved quickly. Hence, in this article, we shall talk about some tricks and strategies that may come in handy to solve questions on quadratic equations quickly and correctly.

Before we get into the tricks, let us be clear that the basic equation in the case of a quadratic equation is ax²+bx+c = 0, where this equation is equal to zero, with a,b,c being a constant number.

In order to better explain, let us consider an example:

6x² + 11x + 3 = 0

Step 1: The first step is to identify the coefficient. Here, the coefficient of x² is 6, x is 11 and the constant number is 3

Step 2: After the coefficient is identified, let us now apply the middle term break method, by applying the following steps:

1. Multiply the coefficient of x² with the constant. In this case, it is (6) X (3) = 18
2. Break the result obtained in previous step into two parts, such that their addition/subtraction results in the middle number, while the multiplication results in the answer obtained in previous step. Hence, in this example, break 18 into two numbers, i.e. 9 and 2. If you notice, the summation of 9 and 2 is 11 (i.e. the coefficient of the middle number), while if we multiple 9 and 2, then we get 18 (i.e. the answer obtained in previous step).
3. The trick to break down the number is by first dividing the result of “a” by 2. If you don’t get a whole number then divide by 3, and so on and forth, until you get a whole number. In this case, divided 18 by 2 first and got the result of 9. Hence the coefficient is 9 and 2.
4. Now change the sign of the factor that you got in “b”, i.e. in this example, change +9 to –9 and +2 to -2. The sign of the coefficient should be change mandatorily.
5. Once the sign is changed, you should now divide both the factors with the coefficient of x². In this example, divide -9 with 6 (i.e. -9/6) and -2 with 6 (i.e. -2/6).  On doing so, you will get the answer -3/2 and -1/3.
6. You may cross check the validity of the equation by replacing x² and x with either -3/2 or -1/3. If you get the result 0, then your answer is correct.

Let us quickly take another example and see if this method can work or not. Below is the equation:

x² + 6x – 9 = 0

Step 1: The coefficient of x² is 1, x is 6 and the constant number is 9

Step 2: Now let us apply the middle term break method, by applying the following steps:

1. Multiply the coefficient of x² with the constant. In this case, it is (1) X (9) = 9
2. Break the result obtained in previous step into two parts, such that their addition/subtraction results in the middle number, while the multiplication results in the answer obtained in previous step. Hence, in this example, break 9 into two numbers, i.e. 3 and 3. If you notice, the summation of 3 and 3 is 6 (i.e. the coefficient of the middle number), while if we multiple 3 and 3, then we get 9 (i.e. the answer obtained in previous step).
3. The trick to break down the number is by first dividing the result of “a” by 2. If you don’t get a whole number then divide by 3, and so on and forth, until you get a whole number. In this case, divided 9 by 2 first but the result was in decimal. Hence, we then divided 9 by 3 and got the result of 3. Hence the coefficient is 3 and 3.
4. Now change the sign of the factor that you got in “b”, i.e. in this example, change +3 to –3 and +3 to -3. The sign of the coefficient should be change mandatorily.
5. Once the sign is changed, you should now divide both the factors with the coefficient of x². In this example, divide -3 with 1 (i.e. -3/1) and -3 with 1 (i.e. -3/1).  On doing so, you will get the answer -3 and -3. Coincidentally the value of x² and x is -3 in both situations.
6. You may cross check the validity of the equation by replacing x² and x with either -3 or -3. If you get the result 0, then your answer is correct.

Solved Example

Now that you have understood how to solve the equation quickly, let us look at how the question is asked in the exam. Here as well, we shall take an example:

Two equations I and II are given. Solve both the equations and give answers by selecting the correct option:

x²– 11x +28 = 0

y² – 18y +81 =0

1. x > y
2. x < y
3. x=y, or relation cannot be established between x and y
4. x ≥ y
5. x ≤ y

Solution:

Now to arrive at the final option, compare both the value of x with both the value of y, as given below:

In all the four combinations, the value of x < y, hence the correct answer is Option B.

Disclaimer:

You may find that there are too many steps involved in solving this question. But if you practise several questions using this method, then it will not even take a minute to solve the question as you will be able to do almost all the steps mentally without using pen and paper. Further, the accuracy rate will be high as well.

General tips:

1. The first and foremost thing you should be doing is to practise as many question as possible, because you can increase your speed only when you practise. For that you may subscribe to mock test series on Quadratic Equation, as through that you will get exposed to variety of questions.
2. Avoid jumping into the options at the first instance as the options will not make any sense until you have solved the value of x and y. Solving quadratic equation questions can be time consuming if followed the traditional method. Hence, that is the reason why we suggest you follow the aforesaid technique.
3. However, in case you are comfortable following the traditional way or are more confident about it then you may follow that.

Quadratic Equations- Questions for Practice

Download the below PDF to access questions for practice.

We hope the above PDF will be helpful in your preparation.

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