# How To Solve Quadratic Equations Quickly

### Solve Quadratic Equations Quickly & Definition

A quadratic equation is an equation having the form ax+ bx + c = 0.
Where,  x is the unknown variable and a, b, c are the numerical coefficients. ## Type 1: Solve Quadratic Equations Questions Quickly

In each of these questions, two equations are given. You have to solve these equations and find out the values and relation of between x and y.

### Question 1

17x2 + 48x – 9 = 0
13y2 – 32y + 12 = 0

Options:

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution

17x2 + 48x – 9 = 0…. (1)

17x2 + 51x – 3x – 9 = 0

(x+3) (17x – 3) = 0

Therefore, Roots of first equation are -3 and 3/17

We know that, if sign given in the equation is + and – then their sign of roots is – and + respectively.

Therefore, the roots of the equation are – 3 and -3/17

Now, 13y2 – 32y + 12 = 0 ……. (2)

13y2 – 26y – 6y + 12 = 0

(y-2) (13y – 6) = 0

Therefore, Roots of second equation are 2 and 6/13

We know that, If sign given in the equation is – and – then their sign of roots is + and + respectively.

Therefore, the roots of the equation are +2 and +6/13

Now, compare the roots -x1, + x2, + y1, and + y2

It means y>x

### Question 2.

√500x – √420 = 0
√260y –
√200= 0

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

√500x – √420 = 0…… (1)

√500x =√420

500x = 420

x = 420/500

x = 210/250

x = 0.84

Now, √260y – √200 = 0

√260y = √200

260y = 200

y = 200/260

y = 100/130

y = 5/9

y = 0.76

On comparing x and y, it is clear that x > y

### Question 3.

x2 – 11x + 24 = 0
2y2– 9y + 9 = 0

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

x2 – 11x + 24 = 0……….. (1)

x2 – 8x – 3x + 24 = 0

(x – 8) (x – 3) = 0

Therefore, Roots of first equation are 8 and 3

We know that, If sign given in the equation is  – and – then their sign of roots is + and +

Therefore, the roots of the equation are +8 and + 3

Now, 2y2  – 9y + 9 = 0 ……. (2)

2y2  – 6y – 3y + 9 = 0

(y-3) (2y – 3) = 0

Therefore, Roots of second equation are 3 and 1.5

We know that, If sign given in the equation is – and + then their sign of roots is + and +

Therefore, the roots of the equation are + 3 and + 1.5

Now, compare the roots +x1, +x2, + y1, and + y2

It means x ≥ y

## Type 2: Quickly Solve Quadratic Equations Questions

When relation cannot be determined

### Question 1.

9x2 – 36x + 35 = 0

2y2 – 15y – 17 = 0

Options

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

9x2 – 36x + 35 = 0……….. (1)

9x2 – 21x – 15x + 35 = 0

(3x – 7) (3x – 5) = 0

Therefore, Roots of first equation are 7/3 and 5/3

We know that, If sign given in the equation is  – and – then their sign of roots is + and + respectively

Therefore, the roots of the equation are +1.66 and +2.33

Now,

2y2 – 15y – 17 = 0……. (2)

2y2 – 17y + 2y – 17 = 0

(y + 1) (2y – 17) = 0

Therefore, Roots of second equation are 8.5 and -1

Now, compare the roots +x1, + x2, + y1, and – y2

It means , we cannot find any relation between x and y

### Question 2.

x2 – 165 = 319

y2 + 49 = 312

Options

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution

x2– 165 = 319……(1)

x2– 165 – 319 = 0

x2= 484  = ±22

y2 + 49 = 312…. (2)

y2 + 49 -312 = 0

y2 = 263 y = ± 16.21

Now compare, x and y It means, we cannot find any relation between x and y

### Question 3.

4x2 + 18x – 10 = 0
y2/5 – 25/y8/5 = 0

Options:

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

### Solution:

4x2 + 18x – 10 = 0……….. (1)

Simplify it, 2x2 + 9x – 5 = 0

2x2 + 10x – x – 5 = 0

(x + 5) (2x – 1) = 0

Therefore, Roots of first equation are 5 and 0.5

We know that, if sign given in the equation is + and – then their sign of roots is – and +

Therefore, the roots of the equation are – 5and +0.5

Now, y2/5 – 25/y8/5 = 0……. (2)

y2/5 – 52/y8/5 = 0

y = ± 5

Now, compare the roots -x1, + x2, ± y

It means, we cannot find any relation between x and y