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Find the natural numbers which differ by 3 and whose squares have the sum 11 |
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Answer» let first no. =xsecond 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 =117x 2 +x 2 +9+6x=1172x 2 +6x=1082(x 2 +3x)=108x 2 +3x=54x 2 +3x−54=0x 2 +9x−6x−54=0x(x+9)−6(x+9)(x−6)(x+9)=0x−6=0x=6 so first no.=6second number =x+3=9please MARK as brainliest |
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