Question

Find two natural numbers differ by 2 and sum of their squares is 100​

Answer 1

Answer:

let first no. =x

second number=x+3

\begin{gathered} {x}^{2} + {(x + 3)}^{2} = 117 \\ {x}^{2} + {x}^{2} + 9 + 6x = 117 \\ 2 {x}^{2} + 6x = 108 \\ 2( {x}^{2} + 3x) = 108 \\ {x}^{2} + 3x = 54 \\ {x}^{2} + 3x - 54 = 0 \\ {x}^{2} + 9x - 6x - 54 = 0 \\ x(x + 9) - 6(x + 9) \\( x - 6)(x + 9) = 0 \\ x - 6 = 0 \\ x = 6\end{gathered}

x

2

+(x+3)

2

=117

x

2

+x

2

+9+6x=117

2x

2

+6x=108

2(x

2

+3x)=108

x

2

+3x=54

x

2

+3x−54=0

x

2

+9x−6x−54=0

x(x+9)−6(x+9)

(x−6)(x+9)=0

x−6=0

x=6

so first no.=6

second number =x+3=9

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