# How To Solve Quadratic Equations Quickly

## How to Solve Simple Interest Questions Quickly in Aptitude

Quadratic equations are actually used in everyday life. It is used to  find areas, product’s profit or express  the speed of an object. Quadratic equations is the equations with at least one squared variable i.e.  ax² + bx + c = 0. Therefore, we consider it as an important topic in the competitive exams and believe that you should know methods related to it and even have a command on How To Solve Quadratic Equation Quickly.

### How To Solve Quadratic Equation & 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.

Example : Solve for x : x2-3x-10 = 0
Solution :  Let us express -3x as a sum of -5x and +2x.

x2-5x+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: Solving 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  Solve for the equations 17x2 + 48x – 9 = 0 and 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 $\frac{3}{17}$

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 $– \frac{3}{17}$

Now, 13y2 – 32y + 12 = 0 ……. (2)

13y2 – 26y – 6y + 12 = 0

(y-2) (13y – 6) = 0

Therefore, Roots of second equation are 2 and $\frac{6}{3}$

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 +2 and $+ \frac{6}{13}$

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

It means y>x

Correct option: A

Question 2. Solve for the equations $\mathbf{\sqrt{500}x =\sqrt{420}}$ and $\mathbf{\sqrt{260}y – \sqrt{200} = 0 }$

Options

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. cannot be determined

Solution:    $\sqrt{500}x -\sqrt{420} = 0$…… (1)

$\sqrt{500}x =\sqrt{420}$

500 x = 420

x =$\frac{420}{500}$

x = $\frac{210}{250}$

x = 0.84

Now, $\sqrt{260}y -\sqrt{200} = 0$

$\sqrt{260}y = \sqrt{200}$

260y = 200

y = $\frac{200}{260}$

y = $\frac{100}{130}$

y = $\frac{5}{9}$

y = 0.76

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

Correct option: B

Question 3. Solve for the equations  $x ^{2} – 11x + 24 = 0 and 2y^{2}– 9y + 9 = 0$

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 +x1, +x2, + y1, and + y2

It means x ≥ y

Correct option: D

### Type 2: How To Solve Quadratic Equation Questions

• When relation cannot be determined

Question 1. Solve for equations 9x2 – 36x + 35 = 0 and 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 $\frac{7}{3}$ and $\frac{5}{3}$

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

Correct option: E

Question 2. Solve for equations  $x^{2} – 165 = 319 and y^{2} + 49 = 312$

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

Correct option: E

Question 3. Solve for Equations   4x2 + 18x – 10 = 0 and  $y^\frac{2}{5}$$(\frac{25}{y})^\frac{8}{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 – 5 and +0.5

Now, $y^\frac{2}{5}$ – $(\frac{25}{y})^\frac{8}{5}$ = 0……. (2)

$y^\frac{2}{5}$  –  $(\frac{5^2}{y})^\frac{8}{5}$ = 0

y = ± 5

Now, compare the roots -x1, + x2, ± y

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

Correct option: E

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