Quadratic equation is one of the complex and tricky parts of Quantitative Aptitude where candidates generally make silly mistakes if they do not focus during preparation. Tricks and strategy are essential for solving a quadratic equation. The **bank exam** requires accuracy in the allotted time during the main exam so shortcut tricks are important for an aspirant to know. The numerical ability section is where the candidates can score well by practicing within a given time using more tricks.

**Weightage of Quadratic Equation in bank exam:**

Priority |
Nature of expected questions |
Weightage |

1 | More accuracy, can be solved easily, easy to manage time, scoring | 0-5 |

**Stepwise tips to practice Quadratic equation:**

- Firstly the candidate needs to note down twenty sums related to quadratic equation on a page.
- Then ten sums are to be solved using basic formula.
- To solve these sums using a stopwatch will help in time management.
- The next step is to evaluate the time taken and analyze the performance.
- The remaining ten questions are to be solved next by applying the shortcut tricks.
- The time taken is to be noted.
- This will surely have a difference from the time taken while solving the first ten questions.
- Practice is the key to master this section.

The basic equation of quadratic equation is **\(ax^{2}+ bx+c =0\)** where this equation is equal to zero and a,b,c are constants. The quadratic equation only holds the power of x where x is also known as a non-negative integer.

**For example**: \(6x^{2} + 11x +3= 0\)

**Solution:** In this equation 6 is coefficient of \(x^{2}\)
11 is the coefficient of x and 3 is a constant. The easiest way is to solve this through middle term break method.

First step:Â To multiply \((+6)\times (+3)\) = +18

Second step: +18 is to be broken into two parts in such a way that their addition result in the middle term that is 11 where \(9 + 2 = 11\) and product of these factors are 18.

Third step: To change the sign of the factors where +9 = -9 and +2 =-2 and this has to be divided by coefficient of \(x^{2}\). Therefore \(-9/6 = -3/2\) and \(-2/6 = -1/3\).

**The major three ways to solve the problems on Quadratic equation are:**

**To factor the quadratic equation-**Here all the same terms are to be combined and to the one side of the equation in such a way that there remains nothing on the other side. Once it has no remaining terms we can write a 0.

**For example**: \(3x^{2} – 8x -4 = 3x + x^{2}\) \(4x^{2} – 11x – 4=0\) The second step is to factor the equation so that there is a set each through middle term break method. The last part is to separate each factor set to zero.

**For example:**\((3x+1)(x-4)=0\)**To use the formula-**The sums can be solved through the formula \(\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\) and solve putting the equations in this.**To complete the square-**This process is a bit complex while the other two are easy to solve.

The above-mentioned shortcut tricks and strategies will help the candidates in preparing for the Quadratic equations. All the candidate needs to do is practice with determination. The **IPBS** and SBI exams both has this section is their two tiers of written examination so candidates are required to practice efficiently.