Question

The sum of the squares of two consecutive
natural numbers is 41. Find the numbers.​

Answer 1

$\\ \\ \underline{ \underline{ \sf \red{given : } }}$

• The sum of the squares of two consecutive natural numbers is 41.

$\\ \\ \underline{ \underline{ \sf \red{to \: find : } }}$

• The two numbers.

$\\ \\ \underline{ \underline{ \sf \red{solution : } }}$

Let the two numbers be (x) and (x+1) as they are consecutive.

Sum of there square is 41.

$\\ \implies\sf \: {x}^{2} + (x + {1)}^{2} = 41$

By identity ,

$\\ \bigstar\boxed{ \bf \: ( {a +b )}^{2} = {a}^{2} + {b}^{2} + 2ab }$

Here ,

• a = x
• b = 1

Puting values ,

$\\ \implies \sf \: {x}^{2} + \{{x}^{2} + {1}^{2} + 2(x)(1) \}= 41 \\ \\ \implies \sf \: {x}^{2} + {x}^{2} + 1 + 2x = 41 \\ \\ \implies \sf \: {2x}^{2} + 2x - 40 = 0 \\ \\ \sf \: dividing \: whole \: equation \: by \:2... \\ \\ \implies \sf \: {x}^{2} + x - 20 = 0$

Now , we will factorise the above equation by Splitting middle term.

$\\ \\ \implies \sf \: {x}^{2} + 5x - 4x - 20 = 0 \\ \\ \implies \sf \: x(x + 5) - 4(x + 5) = 0 \\ \\ \implies \sf \: (x - 4)(x + 5) = 0$

Hence ,

$\\ \\ \sf \: (x -4 ) = 0 \: \: \: \: \: \: or \: \: \: \: \: \: (x + 5) = 0 \\ \\ \boxed{\sf \: x = 4} \: \: \: \: \sf \: or \: \: \: \: \boxed{ \sf \: x = - 5}$

But , according to the given data , numbers are natural. So, they cant be negative.

Hence , x = 4

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Numbers are :-

• x = 4
• x + 1 = 5

Hence , the numbers are 4 and 5.