Question

Find 2 natural numbers which differ by 3 and whose squares have the sum 117

Answer 1

let first no. =x
second number=x+3
${x}^{2} + {(x + 3)}^{2} = 117 \\ {x}^{2} + {x}^{2} + 9 + 6x = 117 \\ 2 {x}^{2} + 6x = 108 \\ 2( {x}^{2} + 3x) = 108 \\ {x}^{2} + 3x = 54 \\ {x}^{2} + 3x - 54 = 0 \\ {x}^{2} + 9x - 6x - 54 = 0 \\ x(x + 9) - 6(x + 9) \\( x - 6)(x + 9) = 0 \\ x - 6 = 0 \\ x = 6$
so first no.=6
second number =x+3=9
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Answer 2

Answer:

Step-by-step explanation:

let first no. =x

second number=x+3

{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

so first no.=6

second number =x+3=9

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