Question

The difference of squares of two natural numbers is 45 . the square of the smaller number is 4 times the larrger number . find the number

Answer 1

Solution is attached below

Answer 2

Answer:

9,6

Step-by-step explanation:

Let the two natural numbers be x and y

Now we are given that The difference of squares of two natural numbers is 45

$x^2-y^2=45$ ---1

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

$y^2=4x$  --2

Substitute value of $y^2$ from 2 in 1

$x^2-4x=45$

$x^2-4x-45=0$

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

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

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

$x=-5,9$

Since we are given that x and y area natural number s.

So, x = 9

So, $y^2=4x=4 \times 9 = 36$

y = 6

So, numbers are 9,6