Question

find the natural number whose sum us 50 product i525​

Answer 1

let the required no. be x and y

let the required no. be x and yso according to the question

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0=> x(x-35) -15(x-35) =0

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0=> x(x-35) -15(x-35) =0=> (x-35)(x-15) = 0

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0=> x(x-35) -15(x-35) =0=> (x-35)(x-15) = 0=> x-35 = 0 or x-15 = 0

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0=> x(x-35) -15(x-35) =0=> (x-35)(x-15) = 0=> x-35 = 0 or x-15 = 0=> x = 35 , 15

let the required no. be x and yso according to the question=> x+ y = 50 -------(1)and => xy = 525 -------(2)so put y = 525/x in eqn (1)we get ,=> x + 525/x = 50=> x^2 + 525 = 50x=> x^2 -50x +525 = 0=> x^2 -35x -15x + 525 =0=> x(x-35) -15(x-35) =0=> (x-35)(x-15) = 0=> x-35 = 0 or x-15 = 0=> x = 35 , 15# so the required number is 35 and 15

Answer 2

Answer:

Let the required no. be x and y

so according to the question

=> x+ y = 50 -------(1)

and => xy = 525 -------(2)

so put y = 525/x in eq (1)

we get ,

=> x + 525/x = 50

=> x^2 + 525 = 50x

=> x^2 -50x +525 = 0

=> x^2 -35x -15x + 525 =0

=> x(x-35) -15(x-35) =0

=> (x-35)(x-15) = 0

=> x-35 = 0 or x-15 = 0

=> x = 35 , 15