Question

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

Answer 1

Hey.

Let the larger no. be x.

so, square of larger no. = x^2

so, the square of smaller no. = 4x

Also, x^2 - 4x = 45

or, x^2 -4x -45 = 0

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

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

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

so, x = 9 .....as -5 will be rejected .

So, the no. is 9 and √4×9 = √36 = 6.

Hence, 9 and 6 are the nos.

Thanks.

Answer 2

solution :

Let x , y are two positive numbers and x > y .

i ) The square of the smaller number is

four times the larger number .

y² = 4x -----( 1 )

ii ) x² - y² = 45 -----( 2 ) [ Given ]

substitute y² = 4x in equation ( 2 ) we get,

x² - 4x = 45

⇒ x² - 4x - 45 = 0

⇒ x² - 9x + 5x - 45 = 0

⇒ x( x - 9 ) + 5 ( x - 9 ) = 0

⇒ ( x - 9 )( x + 5 ) = 0

∴ x - 9 = 0 or x + 5 = 0

x = 9 or x = - 5

But numbers are positive .

∴ x = 9

Substitute x = 9 in equation ( 1 ) we get ,

y² = 4 × 9

⇒ y = √36

y = 6

Required numbers are

x = 9 ,

y = 6

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