Quadratic Equations are the equations of order 2. In the banking exams, these equations form a frequent section of the quantitative aptitude. The quadratic equations are usually easy to factorize and the quadratic formula is seldom used. Here we will see the exact questions and some brief tips and tricks to solve theseÂ quadratic equation questions.

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## Quadratic Equations

The equations of the formÂ ax^{2} + bx + c = 0 or the ones that can be reduced to such form are known ass the quadratic equations. The solutions of this equation are two in number at the maximum and are also known as the roots of the equation. There are two methods that we will discuss here in brief by which we can solve the quadratic equations.

**Browse more Topics under Arithmetic Aptitude**

- Approximations and Simplifications
- Data Sufficiency
- HCF and LCM
- Arithmetic Aptitude Practice Questions

### Factorization

Factorisation, if done correctly will give two linear equations in x. From these equations, we get the value of the variable ‘x’. Letâ€™s see an example and we will get to know more about it.

Example 1: Solve the equation: x^{2} + 3x – 4 = 0

Solution: This method is also known as splitting the middle term method. Here, a = 1, b = 3, c = -4.

The first step is to multiply a and c = 1Â Ã— (-4) = -4. The second step is that the middle term is split into two terms. We do it in such a way that the product of the new coefficients equals the product of a and c that we got above. We have to get 3 here. Therefore, we write x^{2} + 3x – 4 = 0 as x^{2} + 4x – x – 4 = 0. Thus, we can factorise the terms as: (x+4)(x-1) = 0. Using the law that any two quantities a and b, if aÃ—b = 0, we must have either a = 0, b = 0 or a = b = 0.

Either (x+4) = 0 or (x-1) = 0 or both are = 0. This gives x+4 = 0 or x-1 = 0. Solving these equations for x gives: x=-4 or x=1. This method is convenient but is not applicable to every equation. In those cases, we can use the other methods as discussed below.

### Quadratic Equation Formula

There are equations that can’t be reduced using the above two methods. For such equations, a more powerful method is required. A method that will work for every quadratic equation. This is the general quadratic equation formula. We define it as follows: If ax^{2} + bx + c = 0 is a quadratic equation, then the value of x is given by the following formula:

Just plug in the values of a, b and c, and do the calculations. The quantity in the square root is called the discriminant or D. The below image illustrates the best use of a quadratic equation.

Example 2: Solve: x^{2} + 2x + 1 = 0

Solution: Given that a=1, b=2, c=1, and

Discriminant = b^{2} âˆ’ 4ac = 2^{2} âˆ’ 4Ã—1Ã—1 = 0

Using the quadratic formula, x = (âˆ’2 Â± âˆš0)/2 = âˆ’2/2

## Quadratic Equation Questions

The questions are always based on a certain pattern. The following pattern is usually followed:

Example – 1: In each of these questions, two equations (I) and (II) are given. You have to solve both equations and decide which of the following options is correct:

(A) If x < y

(B) If x > y

(C) If x = y or no relationship can be established between x and y.

(D) If x < y

(E) If x > y

Q 1: I: x^{2} + 4x + 4 = 0

II: y^{2} + 3y + 2 = 0

Answer: If you solve the first equation, you will find that x = -2. Let us see how. We can write the first equation as:

x^{2} + 4x + 4 = 0Â â‡’ x(x + 2) + 2(x + 2) = 0

or (x + 2) (x + 2) = 0or x = -2.

From the second equation, we see that y(y + 1) + 2 ( y + 1) = 0Â â‡’ (y + 2) (y + 1) = 0

â‡’ y = -2 and y = -1.

Thus plotting the values of ‘x’ and ‘y’ got from both the equation we see the following:

y | y |

-2 | -1 |

x |

Hence, we say that the actual relation can’t be determined as x may be equal to or less than y which is not any of the options.Â Therefore the correct option here isÂ (C) If x = y or no relationship can be established between x and y.

### Some Other Examples:

Q 2: I: x^{2} + 6x + 9 = 0

II: (âˆš2)y + 1 = 0.

Answer: I: This equation can be solved as follows:

x^{2} + 3x + 3x + 9 = 0Â â‡’ x(x+3) + 3(x+3) = 0

(x + 3) + (x + 3) = 0Â â‡’ x = -3.

II: The second equation can be written as, after transferring 1 to the R.H.S. and squaring both sides, we have:

2y^{2} = 1Â â‡’Â y^{2}Â = 1/2 or y =Â Â± 1/(âˆš2)

Now putting these values on a number line, we have:

-3 | -1/2 | 1/2 |

x | y | y |

So as it is clear from the above description, x is always less than y. Thus the correct option to choose here isÂ (D) x < y.

## Practice Questions

In each of these questions, two equations (I) and (II) are given. You have to solve both equations and decide which of the following options is correct:

(A) If x < y

(B) If x > y

(C) If x = y or no relationship can be established between x and y.

(D) If x < y

(E) If x > y

Q 1: I:Â Â 3x^{2}Â + 2x â€“ 21 = 0

II: 3y^{2}Â â€“ 19y + 28 = 0

Ans: D)Â x < y

Q 2:Â I. 3x^{2}Â + 7x â€“ 6 = 0

II. 4y^{2}Â â€“ 13y â€“ 12 = 0

Ans: C)Â x = y or no relationship can be established between x and y.

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