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