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Answer :
Final answer:
To find two consecutive whole numbers with a sum of squares of 221, set up the equation x^2 + (x+1)^2 = 221, solve the quadratic equation, and find that the consecutive whole numbers are 10 and 11.
Explanation:
To solve this problem, we need to set up two equations. Let's assume the two consecutive whole numbers are x and x+1. The sum of their squares is 221, so we have the equation:
x^2 + (x+1)^2 = 221
Expanding and simplifying the equation gives us:
2x^2 + 2x + 1 = 221
Now, subtract 221 from both sides:
2x^2 + 2x - 220 = 0
We can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. The solutions are x = -11 and x = 10. Since we are looking for whole numbers, the consecutive whole numbers are 10 and 11.
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Rewritten by : Barada
[tex]n^2+(n+1)^2=2n^2+2n+1=221[/tex]
[tex]2n^2+2n=220[/tex]
[tex]n^2+n=110[/tex]
[tex]n^2+n-110=0[/tex]
[tex](n-10)(n+11)=0\implies n=10,n=-11[/tex]
But since [tex]n[/tex] must be a whole number, we ignore [tex]n=-11[/tex]. So the two integers are [tex]n=10[/tex] and [tex]n+1=11[/tex].
[tex]2n^2+2n=220[/tex]
[tex]n^2+n=110[/tex]
[tex]n^2+n-110=0[/tex]
[tex](n-10)(n+11)=0\implies n=10,n=-11[/tex]
But since [tex]n[/tex] must be a whole number, we ignore [tex]n=-11[/tex]. So the two integers are [tex]n=10[/tex] and [tex]n+1=11[/tex].
