PROBLEMS ON QUADRATIC EQUATION

Factor :

Problem 1 :

x² + 6x + 9

Solution :

Product of factors = 9

Sum of factors = 6

Finding the factors of product and sum.

1 · 9 = 9 and 1 + 9 = 10 ≠ 6

3 · 3 = 9 and 3 + 3 = 6

So, the factors are 3 and 3

= x² + 3x + 3x + 9

By grouping,

= (x² + 3x) + (3x + 9)

By taking the common factor, we get

= x(x + 3) + 3(x + 3)

= (x + 3)(x + 3)

Problem 2 :

x² + 3x - 10

Solution :

Product of factors = -10

Sum of factors = 3

Finding the factors of product and sum.

(-1) · 10 = -10 and -1 + 10 = 9 ≠ 3

(-2) · 5 = -10 and -2 + 5 = 3

So, the factors are -2 and 5

= x² - 2x + 5x - 10

By grouping,

= (x² - 2x) + (5x - 10)

By taking the common factor, we get

= x(x - 2) + 5(x - 2)

= (x - 2) (x + 5)

Problem 3 :

What are the roots to the equation x² - 3x - 54 = 0?

a)  x = 6 or x = -9   b)  x = 1 or x = -54 

c)  x = 9 or x = -6  d)  x = 54 or x = -1

Solution :

x² - 3x - 54 = 0

By Splitting the middle term, we get

x² + 6x - 9x - 54 = 0

By taking the common factor, we get

x(x + 6) - 9(x + 6) = 0

(x - 9) (x + 6) = 0

x = 9 or x = -6

So, option (c) is correct.

Problem 4 :

Find the roots of x² - x - 12 = 0

Solution :

x² - x - 12 = 0

By Splitting the middle term, we get

x² + 3x - 4x - 12 = 0

By taking the common factor, we get

x(x + 3) - 4(x + 3) = 0

(x - 4) (x + 3) = 0

x = 4 or x = -3

So, the roots are x = 4 or x = -3.

Problem 5 :

Factor 3x² - x - 14

a)  (3x + 7) (x - 2)   b)  (x + 7) (3x - 2) 

 c)  (3x - 7) (x + 2)  d)  (x - 7) (3x + 2)

Solution :

= 3x² - x - 14

By Splitting the middle term, we get

= 3x² + 6x - 7x - 14

By taking the common factor, we get

= 3x(x + 2) - 7(x + 2)

= (3x - 7) (x + 2)

So, option (c) is correct.

Problem 6 :

Factor the expression 3x² + 15x + 18

Solution :

= 3x² + 15x + 18

Factoring 3,

= 3(x² + 5x + 6)

= 3(x² + 2x + 3x + 6)

By grouping,

= 3[(x² + 2x) + (3x + 6)]

= 3[x(x + 2) + 3(x + 2)]

= 3(x + 2) (x + 3)

Problem 7 :

What are the solutions of the equation m² = - 4m?

Solution :

m² = - 4m

m² + 4m = 0

(m + 0) (m + 4) = 0

m = 0 or m = -4

So, the solution is {0, -4}.

Problem 8 :

A touch tank has a height of 3 feet. Its length is three times its width. The volume of the tank is 270 cubic feet. Find the length and width of the tank.

Solution :

Height = 3 ft

Let l be length, w be the width  and h be the height of the tank.

length = 3w

Volume of cubic tank = 270 cubic feet

length (width) (height) = 270

3w(w) (3) = 270

9w² = 270

w² = 270/9

w² = 30

w = √30

w = 5.47 feet

Problem 9 :

The area A of an equilateral triangle with side length s is given by the formula A = √3/4s². Solve the formula for s. Then approximate the side length of the traffic sign that has an area of 390 square inches

Solution :

A = √3/4s²

Given that, the area of triangle = 390 square inches

390 = √3/4s²

s² = 390(4/√3)

= 900.69

s = √900.69

= 30.01

So, the side length of the square is approximately 30 inches.

Problem 10 :

An in-ground pond has the shape of a rectangular prism. The pond has a depth of 24 inches and a volume of 72,000 cubic inches. The length of the pond is two times its width. Find the length and width of the pond.

Solution :

length = 2(width)

depth = 24 inches

Capacity of pond = 72000

length (width) height = 72000

2w(w) (24) = 72000

w² = 72000/48

w² = 1500

w = √1500

= 38.72 inches

length = 2(38.72)

= 77.45 inches

So, the required length and width are 77.45 inches and 38.72 inches.

Problem 11 :

A person sitting in the top row of the bleachers at a sporting event drops a pair of sunglasses from a height of 24 feet. The function h = −16x² + 24 represents the height h (in feet) of the sunglasses after x seconds. How long does it take the sunglasses to hit the ground?

Solution :

When it reaches the ground the height will become 0.

0 = −16x² + 24

−16x² + 24 = 0

16x² = 24

x² = 24/16

x² = 1.5

x = √1.5

= 1.22 

So, after 1.22 seconds the sunglasses will hit the ground.