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Quadratic equation is an important topic and frequently appears in various competitive exams. A Quadratic equation is written in the form of

**ax**+

^{2}**bx**+

**c**= 0. In this case, a, b and c are constants, x is a variable and the value of ‘a’ cannot be zero. The presence of

**x**in the equation makes it quadratic else it will be called a linear equation.

^{2}The number of questions asked from this topic range from 3 to 5. The exams where it is asked frequently include SBI PO, IBPS PO, IBPS RRB, IBPS Clerk, SBI Clerk, NIACL, LIC AAO etc. Therefore, candidates need to practice enough number of questions to develop an understanding of the topic.

##

Quadratic Equation Questions PDF:

### Quadratic Equation PDF Set 1

### Quadratic Equation PDF Set 2

Quadratic equation can often be tricky and is generally where candidates make silly mistakes. Most questions from the topic are asked in the first phase of the exam and are also helpful in various other word problems which can be solved by forming equations.

## Solve Quadratic Equation Questions by Factorisation

Quadratic equations can be solved by using more than one method. In this section, you will learn how to factorise quadratic equations and to find its roots. It would be helpful if you keep in mind that to factorise quadratic polynomials, you need to split the middle term. You also should remember that to determine the roots, you need to factorise the equation into linear factors and then equate each factor to zero. After solving the quadratic equation and finding its roots, you must verify that these the roots of the equation are given.

## Solve Quadratic Equation Questions by Methods of Square

To solve a quadratic equation and find its roots, under this method, the equation ax

^{2}+ bx + c = 0 is converted into the form of (x + a)

^{2}– b

^{2}= 0. Then, you have to apply your knowledge of square and square roots to solve the equation. Hence, the method is known as the method of completing the square. The terms which contain x is completely inside a square and the roots were found via taking the square roots.

## The Nature of Roots

A quadratic equation ax

^{2}+ bx + c = 0 has

1) Two distinct real roots, when b

^{2}– 4ac > 0.

2) Two equal real roots, when b

^{2}– 4ac = 0.

3) No real roots, when b

^{2}– 4ac < 0.

A clear grasp of this concept will help you to understand more about the nature of the roots.

## Tips to solve Quadratic Equation Questions

The first step is to note down five sums related to the quadratic equation on a page. Then try to solve five sums using a basic formula. Solve these sums using a stopwatch for better time management. The next step is to evaluate the time taken and then analyse your performance.

The try solving the questions by applying shortcut tricks and note down the time carefully. It will surely make a difference from the time when you solved the first set of questions. Practice is the key to master these questions, so attempt the questions consistently.

The basic equation is ax2+bx+c=0 where this equation is equal to zero and a,b,c are constants. The quadratic equation holds the power of x where x is known as a non-negative integer.

For example:

**In this question two equations (I) and (II) are given. You have to solve both the equations and give answer.**

**I.**8x

^{2}– 22x + 12 = 0

**II.**15y

^{2}– 19y + 6 = 0

A. if x > y

B. if x ≤ y

C. if x ≥ y

D. if x < y

E. if x = y or relationship between x and y can't be established

**Ans. A**

**I.**8x

^{2}– 22x + 12 = 0

⇒ 8x

^{2}– 16x – 6x + 12 = 0

⇒ 8x (x – 2) – 6 (x – 2) = 0

⇒ (8x – 6) (x – 2) = 0

⇒ x = |
6 |
= 3/4, 2 |

8 |

**II.**15y

^{2}– 19y + 6 = 0

⇒ 15y

^{2}– 10y – 9y + 6 = 0

⇒ 5y (3y – 2) – 3 (3y – 2) = 0

⇒ (5y – 3) (3y – 2) = 0

⇒ y = |
3 |
. |
2 |

5 | 3 |

While comparing the root values of x and y, we find that both the values of x are greater than y's.

Hence, option A is correct.

**Tips to solve Quadratic Equations**

Remember the major ways to solve the problems on the Quadratic Equation:

1. Factor the Quadratic Equation- In this case, all the same terms are to be combined to the one side of the equation in such a way that there is nothing on the other side. The second step will be to factor the equation in order to set each through the middle term break method. The last part will be to separate each factor set to zero.

2. Using the formula- The sum can be solved using the formula − b ± b

^{2}−4ac√2a and solve putting the equations in this formula.