# Tips And Tricks And Shortcuts For Quadratic Equations

## Tips and Tricks and Shortcuts for Quadratic Equation

Quadratic Equation is an equation which is written in the standard form is ax² + bx + c = 0 with a, b, and c being constants and x is an variable. That’s why it is also known as second degree equation . Here, we will be discussing about tips and tricks for quadratic equations.

• 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 competitive exams and recruitment exams.
• There are generally two in of Quadratic Equations.
1. Factorization
2. Completing the square method.

### Tips and shortcuts 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: x2 + 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: 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. Find the correct option given below by solving the following equation  x2 – 7x + 10 = 0 and y2 + 8y + 15 = 0

Options:

A. x < y
B. x > y
C. x ≤ y
D. x ≥ y
E. cannot be determined

Solution:     x2 – 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, y2 + 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 Find the correct option given below by solving the following equation 6x2 + 11x – 35 = 0 and  6y2 + 5y – 6 = 0

Options:

A. x < y
B. x > y
C. x ≤ y
D. x ≥ y
E. cannot be determined

Solution:     6x2 + 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, 6y2 + 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 -x1, + x2, – y1, and + y2

It means x2> – y1 and –x1< + y2

So, we cannot determine which roots are greater.

Correct option: E

### Type 3: Tips and Tricks and Shortcuts for Quadratic Questions

• When we can find a equation where x is a variable and a, b, and c represent constants and D, i.e., Discriminant.
• The discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac.

Question 1 Find the Discriminant and roots of the following equation
3x2 − 5x − 7 = 0

Let us find Discriminant, D = $\sqrt{B^{2}-4AC}$

D = $\sqrt{(-5)^{2}-4(3)(-7)}$

D = 109

Therefore, $x = \frac{-b \pm \sqrt{}D}{2a}$

x = $\frac{5 \pm \sqrt{}109}{6}$

x = $\frac{5 \pm 10.44}{6}$

x =$\frac{15.44}{6} => x = 2.57$, and

x = $\frac{-5.44}{6} => x = – 0.90$

So the two roots are 2.57 and -0.90

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