Question

Five times of a positive integer is less than twice its square by 3.
Find the integer.

Answer 1

Answer :

3

Solution :

Let the required positive integer be x .

Now ,

According to the question , five times of the positive integer is less than twice its square by 3 .

Thus ,

=> 2x² - 5x = 3

=> 2x² - 5x - 3 = 0

=> 2x² - 6x + x - 3 = 0

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

=> (x - 3)(2x + 1) = 0

=> x = 3 , -1/2

=> x = 3 (appropriate value)

[ Note : x = -1/2 is rejected value because it is not a positive integer ]

Hence ,

Required integer is 3 .

Answer 2

Answer:

Given :

• Five times of a positive integer is less than twice its square by 3.

To Find :

• Find the integer.

Solution :

• Let the number be x,

• 5 times of x =  5x

• 3 less than twice of  square of x = 2x² - 3

According to the Question :

So, our equation is :

2x² - 5x = 3

2x² - 5x - 3 = 0 (Using splitting method)

2x² + x - 6x - 3

x (2x + 1) - 3 (2x + 1)

= (2x + 1) (x - 3)

2x + 1 = 0

2x = - 1

x = - 1/2

x - 3 = 0

x = 3

x can't be negative 

So, THEREFORE, THE NUMBER IS 3.