# Tips And Tricks And Shortcuts For Quadratic Equations

## Quadratic Equation- Tips and Tricks & Shortcuts

On, solving the quadratic equations generally we get two linear equations. Here, are some quick shortcuts and easy tips and tricks for you to solve Quadratic Equation questions quickly, easily, and efficiently in various exams of recruitment drive and competitive exams and recruitment exams.

- There are generally two in of Quadratic Equations.
- Factorization
- Completing the square method.

### Tips and tricks to solve the Quadratic Equation questions

• In order to find the sign of roots we use sign given in the equation.

Sign of coefficient ‘x’ | Sign of coefficient ‘y’ | Sign of roots |

+ | + | – – |

+ | – | – + |

– | + | + + |

– | – | + – |

For example: x^{2} + 3x – 4 = 0

The factors are 4 and 1

Now will find the sign of roots. According to the table, if sign given in the equation is + and – then their sign of roots is – and +.

Therefore, the roots of the equation are -4 and 1

## Type 1: Tips and Tricks and Shortcuts for Quadratic Questions

When one equation has positive roots and other has negative, or equations has same sign roots, then compare the roots and find the relation between them.

### Question 1

x^{2} – 7x + 10 = 0

y^{2} + 8y + 15 = 0

Options:

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

x^{2} – 7x + 10 = 0…. (1)

Roots of first equation are 5 and 2

Using the table above, we will find the sign of roots. If sign given in the equation is – and + then their sign of roots is + and +

Therefore, the roots of the equation are 5 and 2

Now, y^{2} + 8y + 15 = 0 ……. (2)

Roots of second equation are 3 and 5

If sign given in the equation is + and + then their sign of roots is – and –

Therefore, the roots of the equation are -3 and – 2

Now, here roots are +x1, + x2, – y1, and – y2

This clearly shows that X roots are positive and Y roots are negative. So, x> y

#### Correct option: B

## Type 2: Tips and Tricks To Solve Quadratic Equations Questions

When roots of the equation are positive and negative and we cannot find the relation between them.

### Question 1

6x^{2} + 11x – 35 = 0

6y^{2} + 5y – 6 = 0

Options:

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

6×2 + 11x – 35 = 0…. (1)

Roots of first equation are 3.5 and 1.66

Using the table above, we will find the sign of roots. If sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 3.5 and 1.66

Now, 6y^{2} + 5y – 6 = 0 ……. (2)

Roots of second equation are 1.5 and 0.66

If sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 1.5 and 0.66

Now, here roots are -x_{1}, + x_{2}, – y_{1}, and + y_{2}

It means x_{2}> – y_{1} and –x_{1}< + y_{2}

So, we cannot determine which roots are greater.

Please Login/Signup to comment