# How To Solve Quadratic Equations Quickly

### Solve Quadratic Equations Quickly & Definition

A quadratic equation is an equation having the form ax^{2 }+ bx + c = 0.

Where, x is the unknown variable and a, b, c are the numerical coefficients.

**Read Also**–**Formulas to solve quadratic equation questions easily**For example -Solve for x: x

^{2}-3x-10 = 0

SolutionLet us express -3x as a sum of -5x and +2x.

→ x^{2}-5x+2x-10 = 0

→ x(x-5)+2(x-5) = 0

→ (x-5)(x+2) = 0

→ x-5 = 0 or x+2 = 0

→ x = 5 or x = -2

## 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

17x^{2} + 48x – 9 = 0

13y^{2} – 32y + 12 = 0

**Options:**

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

**Solution**

17x^{2} + 48x – 9 = 0…. (1)

17x^{2} + 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, 13y^{2} – 32y + 12 = 0 ……. (2)

13y^{2} – 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, + x_{2}, + y_{1}, and + y_{2}

It means y>x

#### Correct option: A

### 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

#### Correct option: B

### Question 3.

x^{2} – 11x + 24 = 0

2y^{2}– 9y + 9 = 0

**Options**

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

**x ^{2} **– 11x + 24 = 0……….. (1)

**x ^{2} ** – 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, 2y** ^{2} ** – 9y + 9 = 0 ……. (2)

2y** ^{2} ** – 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 +x_{1}, +x_{2}, + y_{1}, and + y_{2}

It means x ≥ y

#### Correct option: D

## Type 2: Quickly Solve Quadratic Equations Questions

When relation cannot be determined

### Question 1.

** 9x ^{2} – 36x + 35 = 0 **

**2y ^{2} – 15y – 17 = 0**

**Options**

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

#### Solution:

9x^{2} – 36x + 35 = 0……….. (1)

9x^{2} – 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,

2y^{2} – 15y – 17 = 0……. (2)

2y^{2} – 17y + 2y – 17 = 0

(y + 1) (2y – 17) = 0

Therefore, Roots of second equation are 8.5 and -1

Now, compare the roots +x_{1}, + x_{2}, + y_{1}, and – y_{2}

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

#### Correct option: E

### 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

x^{2}– 165 = 319……(1)

x^{2}– 165 – 319 = 0

x^{2}= 484 = ±22

y^{2} + 49 = 312…. (2)

y^{2} + 49 -312 = 0

y^{2} = 263 y = ± 16.21

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

#### Correct option: E

### Question 3.

4x^{2} + 18x – 10 = 0

y^{2/5} – 25/y^{8/5} = 0**Options:**

A. x < y B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

### Solution:

4x^{2} + 18x – 10 = 0……….. (1)

Simplify it, 2x^{2} + 9x – 5 = 0

2x^{2} + 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, y^{2/5} – 25/y^{8/5} = 0……. (2)

y^{2/5} – 5^{2}/y^{8/5} = 0

y = ± 5

Now, compare the roots -x_{1}, + x_{2}, ± y

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

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