PROBLEMS ON SOLVING QUADRATIC EQUATION

Solve the equation using any convenient method.

Problem 1 :

x² = 64

Solution:

x² = 64

Using square root property, take square root on both sides, we get

x = √64

x = ±8

x = 8 or x = -8

Problem 2 :

(x - 5)² = 8

Solution:

(x - 5)² = 8

x² + 25 - 10x = 8

x² - 10x + 25 - 8 = 0

x² - 10x + 17 = 0

Use the quadratic formula,

Problem 3 :

4x² - 12x + 9 = 0

Solution:

4x² - 12x + 9 = 0

Using factoring, 

4x² - 6x - 6x + 9 = 0

(4x² - 6x) - (6x - 9) = 0

2x(2x - 3) - 3(2x - 3) = 0

(2x - 3)(2x - 3) = 0

2x - 3 = 0

2x = 3

x = 3/2

Problem 4 :

9x² + 12x + 3 = 0

Solution:

9x² + 12x + 3 = 0

Using factoring,

9x² + 9x + 3x + 3 = 0

9x(x + 1) + 3(x + 1) = 0

(9x + 3) (x + 1) = 0

9x + 3 = 0 or x + 1 = 0

9x = -3 or x = -1

x = -1/3 or x = -1

Problem 5 : 

x² - 6x + 4 = 0

Solution:

x² - 6x + 4 = 0

Use the quadratic formula,

Problem 6 :

2x² - 4x - 6 = 0

Solution:

2x² - 4x - 6 = 0

2x2 + 2x - 6x - 6 = 0

2x(x + 1) - 6(x + 1) = 0

(2x - 6)(x + 1) = 0

2x - 6 = 0 or x + 1 = 0

2x = 6 or x = -1

x = 3 or x = -1

Problem 7 :

50 + 5x = 3x²

Solution:

50 + 5x = 3x²

3x² - 5x - 50 = 0

3x2 - 15x + 10x - 50 = 0

3x(x - 5) + 10(x - 5) = 0

(3x + 10)(x - 5) = 0

3x + 10 = 0 or x - 5 = 0

3x = -10 or x = 5

x = -10/3 or x = 5

Problem 8 :

2x² + 4x - 9 = 2(x - 1)²

Solution:

2x² + 4x - 9 = 2(x - 1)²

2x² + 4x - 9 = 2(x² + 1 - 2x)

2x² + 4x - 9 = 2x² + 2 - 4x

2x² - 2x² + 4x + 4x - 9 - 2 = 0

8x - 11 = 0

8x = 11

x = 11/8

Problem 9 :

Find two consecutive positive even integers whose product is 288. 

Solution:

Let the two consecutive positive integer be x and x+2.

x(x + 2) = 288

x² + 2x = 288

x² + 18x - 16x - 288 = 0

x(x + 18) - 16(x + 18) = 0

(x - 16)(x + 18) = 0

x - 16 = 0 or x + 18 = 0

x = 16 or x = -18

Since the integers are positive.

We wil take x = 16

Therefore another positive integer will be 16 + 2 = 18.

Hence the positive integers are 16 and 18.

Problem 10 :

A triangular sign has a height that is twice its base. The area of the sign is 10 square feet. Find the base and height of the sign.

Solution:

Given, area = 10 square feet

Base = x

height = 2x

Area of triangle = (1/2)bh

10 = 1/2(x)(2x)

10 = x²

base x = √10 feet 

height = 2√10 feet

So, base and height is √10 feet and 2√10 feet.