Question

The sum of squares of two natural numbers is 34 . lf the first number i s 1 less twice than the second number, find the numbers.

Answer 1

Let the second number be a

First number = 2a - 1

According to question,

( 2a - 1)^2 + (a)^2 = 34

=> 4a^2 - 4a + 1 + a^2 = 34

=> 5a^2 - 4a - 33 = 0

=> 5a^2 - 15a + 11a - 33 = 0

=> 5a ( a - 3) + 11(a - 3) = 0

=> ( a - 3) (5a + 11) = 0

a = 3 and - 11 /5

Since it is a natural number, it can't be negative.

a = 3

First number = 2 (3) - 1 = 6 - 1 = 5

Second number = 3

Answer 2

Hey .....!!

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Let the two numbers be x and y !

According to the question !

${x}^{2} + {y}^{2} = 34 \\ \\ x = 2y - 1 \\ \\ By\: substitution\: method \\ \\ {(2y - 1)}^{2} + {y}^{2} = 34 \\ \\ 4{y}^{2} + {y}^{2} - 4y + 1 = 34 \\ \\ 5{y}^{2} - 4y + 1 = 34 \\ \\ 5{y}^{2} - 4y - 33 = 0 \\ \\ 5{y}^{2} - 15y + 11y - 33 = 0 \\ \\ 5y ( y - 3 ) + 11 ( y - 3 ) \\ \\ 5y + 11 = 0 , y - 3 = 0 \\ \\ y = \frac{-11}{5} , y = 3 \\ \\ Neglecting\: negative \:value \\ \\ Putting \: value\: of \:y \\ \\ x = 2y - 1 \\ \\ x = 2(3) - 1 \\ \\ x = 5 \\ \\ First\: number = 5 \:Second\: number = 3$

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