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# 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**^{2 }+ bx + c = 0.

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

**Example** : Solve for x : x^{2}-3x-10 = 0**Solution : **Let 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: 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 17x ^{2} + 48x – 9 = 0 and 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 \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, 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 \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, + x_{2}, + y_{1}, and + y_{2}

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 +x_{1}, +x_{2}, + y_{1}, and + y_{2}

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 9x ^{2} – 36x + 35 = 0 and 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 \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,

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. 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 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. Solve for Equations 4x^{2} + 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:** 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 – 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 -x_{1}, + x_{2}, ± y

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

**Correct option: E**

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