Almost all the students feel difficulty in solving quadratic equations. Don’t worry we are here to help you overcome the difficulties in solving quadratic equations in a simple method. An algebraic expression with degree two is called a quadratic equation. It is a univariable expression with x. Solve the quadratic equation by substituting in the standard form of the equation. In this article, the students can find how to solve quadratic equations with examples of quadratic equations provided in different methods with step-by-step explanations.

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## Quadratic Equations Examples with Answers

Students who wish to score the highest marks in the exams are suggested to go through the below examples of quadratic equations. Try to solve the given problems in different methods so that you can understand the concept and prepare the questions on your own.

**Example 1.**

Solve the equation x² + x = 0 by factoring.

**Solution:**

Given the equation x² + x = 0

Substitute the given equation in the standard form.

ax² + bx + c = 0.

x(x + 1) = 0

x = 0 or x + 1 = 0

x + 1 = 0

x = -1

Thus the solution is {0, -1}

**Example 2.**

Solve the equation x² – 12x + 11 = 0

**Solution:**

Given the equation,

Substitute the given equation in the standard form.

ax² + bx + c = 0.

x² – 12x + 11 = 0

x² – 11x – 1x + 11 = 0

x(x – 11) -1 (x – 11) = 0

(x – 1) (x – 11) = 0

x – 1 = 0

x = 1

x – 11 = 0

x = 11

Thus the solution is {1, 11}

**Example 3.**

Find the range of function f(x) = (x+2)/(2x² + 3x + 6), if x is real.

**Solution:
**Given the equation,

f(x) = (x+2)/(2x² + 3x + 6)

Substitute the given equation in the standard form.

ax² + bx + c = 0.

Let y = 2x² + 3x + 6

2x²y + 3xy + 6y = x + 2

2x²y + (3y – 1)x + 6y – 2 = x + 2

Now discriminant (3y – 1)² – 4(2y)(6y – 2) ≥ 0

9y² + 1 – 6y – 48y² + 16y ≥ 0

39y²- 10y – 1 ≤ 0

39y²- 13y + 3y – 1 ≤ 0

13y(3y – 1) + 1(3y – 1) ≤ 0

Thus the range of quadratic equation f(x) = (x+2)/(2x² + 3x + 6)

y ∈ [-1/13, 1/3]

**Example 4.**

Solve the quadratic equation 2x² + 3x – 2 = 0

**Solution:**

Given the equation,

2x² + 3x – 2 = 0

Substitute the given equation in the standard form.

ax² + bx + c = 0.

a = 2, b = 3, c = -2

x = [-b ± √(b² – 4ac)]/2a

x = [-3 ± √(3² – 4(2)(-2))]/2(2)

x = [-3 ± √(16)]/4

x = [-3 ± 4]/4

x = [-3 + 4]/4 = 1/4 = 0.25

x = [-3 – 4]/4 = -7/4

Thus the solution set is {1/4, -7/4}

**Example 5.**

Solve the quadratic equation x² + 10x + 9 = 0

**Solution:**

Given the equation,

Substitute the given equation in the standard form.

ax² + bx + c = 0.

a = 1, b = 10, c = 9

x = [-b ± √(b² – 4ac)]/2a

x = [-10 ± √(10² – 4(1)(9))]/2(1)

x = [-10 ± √(100 – 36)]/2

x = [-10 ± √64]/2

x = [-10 ± 8]/2

x = [-10 + 8]/2 = -2/2 = -1

x = -1

x = [-10 – 8]/2

x = [-18]/2

x = -9

Thus the solution set is {-1, -9}

**Example 6.**

Solve the quadratic equation x² + 4x + 4 = 0

**Solution:**

Given the equation,

x² + 4x + 4 = 0

Substitute the given equation in the standard form.

ax² + bx + c = 0.

a = 1, b = 4, c = 0

x² + 2x + 2x + 4 = 0

x(x + 2) + 2(x + 2) = 0

(x + 2)(x + 2) = 0

x + 2 = 0

x = -2

x + 2 = 0

x = -2

The roots of the two solutions of the equation are the same.

**Example 7.**

The product of two odd consecutive integers is 48. Find the integer.

**Solution:**

Given,

The product of two odd consecutive integers is 48

Let the two odd consecutive integers be x, x + 2

x(x + 2) = 48

x² + 2x = 48

x² + 2x – 48 = 0

Substitute the given equation in the standard form.

ax² + bx + c = 0

x² + 2x – 48 = 0

x² + 8x – 6x – 48 = 0

x(x + 8) -6(x + 8) = 0

x + 8 = 0

x = -8

x – 6 = 0

x = 6

Thus the solutions are {-8, 6}

**Example 8.**

The product of two consecutive even integers is 24. Find the equation.

**Solution:**

Given,

The product of two consecutive even integers is 24.

Let the two consecutive even integers be (x + 2), x + 4

(x + 2)(x + 4) = 24

x(x + 4) + 2(x + 4) = 24

x² + 4x + 2x + 8 = 24

x² + 6x + 8 = 24

x² + 6x – 16 = 0

**Example 9.**

Solve the quadratic equation x² + 2x + 1 = 0

**Solution:**

Given the equation,

Substitute the given equation in the standard form.

ax² + bx + c = 0.

a = 1, b = 2, c = 1

x² + 2x + 1 = 0

x² + 1x + 1x + 1 = 0

x(x + 1) + 1(x + 1) = 0

(x + 1)(x + 1) = 0

x + 1 = 0

x = -1

x + 1 = 0

x = -1

Thus the roots of the equation are {-1, -1}

**Example 10.**

Solve the quadratic equation 25x² – 16 = 0

**Solution:**

Given the equation,

25x² – 16 = 0

Substitute the given equation in the standard form.

ax² + bx + c = 0.

25x² – 16 = 0

We can solve by squaring the equation.

25x² = 16

Applying square root on both sides.

√25x² = √16

5x = 4

5x – 4 = 0

Squaring on both sides

(5x – 4)² = 0

(5x + 4)(5x – 4) = 0

5x + 4 = 0

5x = -4

x = -4/5

5x – 4 = 0

5x = 4

x = 4/5

Thus the solution is {-4/5, 4/5}