Question

the sum of the squares of two consecutive natural number is 41 . find rhe number

Answer 1

let the number be x other no. is x+1
atq,
x^2 + (x+1)^2=41
now break the (x+1)^2 in (a+b)^2
and then do the middle term factorization and get the answer.
IF IT HEPLS THEN PLZ MARK AS BRAINLIEST.

Answer 2

Answer:

Let the number be x and x + 1.

__________________....

$\underline{ \large \purple{ \mathscr{\dag\:A \bf{ccording} \: to \: \mathscr {Q} \bf{uestion} ....}}}$

$:\implies\sf x^2 + (x + 1)^2 = 41 \\\\\\:\implies\sf 2x^2 + 2x + 1 - 41 = 0\\\\\\:\implies\sf 2x^2 + 2x + 40 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + x - 20 = 0\\\\\\:\implies\sf x^2 + 5x - 4x - 20 = 0\\\\\\:\implies\sf x(x + 5) - 4(x + 5) = 0\\\\\\:\implies\sf (x + 5)(x - 4) = 0\\\\\\:\implies\underline{\boxed{\sf x = -5, 4}}$

But, as we know -5 is not a natural number,

Hence, x = 4 .

So the 2 unknown numbers are :

• x = 4 [1st number]
• x + 1 = 4 + 1 = 5 [2nd number]