Question

The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

Answer 1

Given :-

• The difference of squares of two natural numbers is 45

• The square of the smaller number is 4 times the larger number.

Solution :-

Let the two natural numbers be x and y ,

Here, x > y

According to the question,

x^2 + y^2 = 45. eq( 1 )

Now,

y^2 = 4x. eq( 2 )

Subsitute eq( 2 ) in eq( 1 )

x^2 - 4x = 45

x^2 - 4x - 45 = 0

[ Now, it is a quadratic equation ]

By factorization method,

x^2 - 9x + 5x - 45 = 0

x( x - 9) + 5( x - 9 ) = 0

( x + 5 ) ( x - 9 ) = 0

We get two factors :-

x = - 5 , x = 9

[ Natural numbers never be negative ]

Therefore,

x = 9

Now, Subsitute value of x in eq( 2)

y^2 = 4x

y^2 = 4 * 9

y^2 = 36

y = √36 = √ 6 * 6 = 6

Hence, The two natural numbers are 9 and 6 $.$

Answer 2

Answer:

let the larger number is:- x

smaller numbers is :- 4x

according to the question

x-4x =45

3x= 45

x=45\3

x= 15

Larger number = 15

smaller number= 60