Math, asked by uuu60, 11 months ago

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

Answers

Answered by QueenOfKnowledge
13

Answer:

Let the smaller natural number be x and larger natural number be y

Hence x2 = 4y → (1)

Given y2 – x2 = 45

⇒ y2 – 4y = 45

⇒ y2 – 4y – 45 = 0

⇒ y2 – 9y + 5y – 45 = 0

⇒ y(y – 9) + 5(y – 9) = 0

⇒ (y – 9)(y + 5) = 0

⇒ (y – 9)= 0 or (y + 5) = 0

∴ y = 9 or y = -5

But y is natural number, hence y ≠ - 5

Therefore, y = 9

Equation (1) becomes,

x2 = 4(9) = 36

∴ x = 6

Thus the two natural numbers are 6 and 9.

Answered by RaviMKumar
5

Answer:

9,6

Step-by-step explanation:

let a,b be the two natural no.s

=> a² - b² = 45 and          ------------------------- 1

   b² = 4a ,  -------------------------------------------- 2

find a,b ?

substitute 2 in 1 =>

a² - 4a = 45

a²-4a-45 = 0

on factorising , we get (a-9)(a+5) =0

                                    so, either a-9 = 0 => a=9

                                           or       a+5 = 0 => a=-5

eliminate a=-5 as square of a no. cannot be negative.(see 2 )

so a = 9

and by 2, b² = 4*9

            => b² = 36

            => b = 6

                                 

 

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