Heatboy62 Heatboy62
  • 09-10-2018
  • Mathematics
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Find two consecutive positive integers such that the sum of their squares is 841

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chisnau chisnau
  • 18-10-2018

Let x = 1st positive integer

Let x + 1 = 2nd positive integer

As give, their sum of squares is 841

[tex]x^{2}+(x+1)^{2}=841[/tex]

[tex]x^{2}+x^{2}+2x+1=841[/tex]

[tex]2x^{2}+2x+1=841[/tex]

[tex]2x^{2}+2x-840=0[/tex]

factor 2 out, [tex]x^{2}+x-420=0[/tex]

[tex]x^{2}+21x-20x-420=0[/tex]

[tex]x(x+21)-20(x+21)=0[/tex]

(x+21)(x-20)

This gives, x=20 and =-21 (neglect the negative value)

So, x = 20

x+1 = 21

So, the two numbers are : 20 and 21



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