Question

a natural number is greater than other by 5.the sum of their squares is 73.find the numbers

Answer 1

Let one number be X

Other number be Y

X - Y = 5

Y = X - 5

Given,

X^2 + Y^2 = 73

X^2 + (X-5)^2 = 73

X^2 + X^2 + 25 - 10X = 73

2X^2 - 10X + 25 - 73 = 0

2X^2 - 10X - 48 = 0

X^2 - 5X - 24 = 0

X^2 - 8X + 3X - 24 = 0

X(X-8) + 3(X-8) = 0

(X+3)(X-8) = 0

So, X = -3 or 8

Since its a natural number

X = 8

Y = 8 - 5 = 3

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Answer 2

Let the 2 natural numbers be x & y.
A. T. Q,
x = y + 5
And also
$\: \: \: \: \: \: {x}^{2} + {y}^{2} = 73 \\ = > {(y + 5)}^{2} + {y}^{2} = 73 \\ = > 2 {y}^{2} + 10y + 25 - 73 = 0 \\ = > 2 {y}^{2} + 10y - 48 = 0 \\ = > {y}^{2} + 5y - 24 = 0 \\ = > {y}^{2} + 8y - 3y - 24 = 0 \\ = > y(y + 8) - 3(y + 8) = 0 \\ = > (y + 8)(y - 3) = 0 \\ = > y = ( - 8)and(3).$

After solving this equation
Put the value of y (which will come) in the first equation and find x.

Then,
The numbers are 3 & 5(Neglecting negative value of y as they are given natural numbers).


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